Answer:
a + s = 394 total tickets sold is 394 s = a - 56 56 fewer student tickets than adult substitute a-56 in place of s in 1st equation and solve for a a + (a - 56) = 394 2a - 56 = 394 2a = 450 a = 225 adult tickets sold s = a-56 = 169 student tickets sold
Hoped I helped
Find the velocity, acceleration, and speed of a particle with position function r(t)=⟨−8tsint,−8tcost,2t2⟩
Answer:
The answer is below
Step-by-step explanation:
Velocity is the rate of change of displacement. Velocity is the ratio of distance to time.
The velocity v(t) = [tex]\frac{d}{dt}r(t)[/tex]
Where r(t) is the position function
Given that:
r(t)=⟨−8tsint,−8tcost,2t²⟩
[tex]v(t)=\frac{d}{dt}r(t)= <-8tcost-8sint,8tsint-8cost,4t>[/tex]
Acceleration is the rate of change of velocity, it is the ratio of velocity to time. Acceleration a(t) is given as:
[tex]a(t)=\frac{d}{dt}v(t)= \frac{d}{dt} <-8tcost-8sint,8tsint-8cost,4t>\\=<8tsint-16cost,8tcost+16cost,4>\\\\a(t)=<8tsint-16cost,8tcost+16cost,4>[/tex]
Speed = |v(t)| = [tex]\sqrt{(-8tcost-8sint)^2+(8tsint-8cost)^2+(4t)^2}\\\\ =\sqrt{64t^2cos^2t+128tcostsint+64sin^2t+64t^2sin^2t-128tsintcost+64cos^2t+16t^2}\\ \\=\sqrt{64t^2cos^2t+64t^2sin^2t+64sin^2t+64cos^2t+16t^2}\\\\=\sqrt{64t^2(cos^2t+sin^2t)+64(sin^2t+cos^2t)+16t^2}\\\\=\sqrt{64t^2+64+16t^2}=\sqrt{80t^2+64}[/tex]
Number of times the individual changed jobs in the last 5 years is what kind of variable? A. This variable is a continuous numerical variable that is interval-scaled. B. This variable is a discrete numerical variable that is interval-scaled. C. This variable is a categorical variable that is ordinal-scaled. D. This variable is a discrete numerical variable that is ratio-scaled. E. This variable is a continuous numerical variable that is ratio-scaled. F. This variable is a categorical variable that is nominal-scaled.
Answer: D. This variable is a discrete numerical variable that is ratio-scaled.
Step-by-step explanation:
A Discrete variables are variables which are countable in a finite amount of time. For example, you can count the amount of money in your bank wallet, but same can’t be said for the money you have deposited in eveyones bank account as this is infinite.
So the number of times an individual changes job in a five years period is a perfect example of a discrete numerical variable that is ratio scaled because it can be counted.
which fraction is less then 7/10
Answer:
Well, there is a lot of answers to that. Check explanation, please!
Step-by-step explanation:
For example, some basic fractions that are less than 1/7 are:
1/8 (Compare their size!)2/18Or 1/154You can also set up a number line, and compare the fractions.
Hopefully, this answer helps! :D
Find an equation of the line passing through the point (−3,−7) that is perpendicular to the line y= −5x+4
Answer:
y = 1/5x - 6.4
Step-by-step explanation:
Perpendicular lines have opposite reciprocal slopes, so the slope will be 1/5
Then, plug the slope and the point into the equation y = mx + b to find b
y = mx + b
-7 = 1/5(-3) + b
-7 = -0.6 + b
-6.4 = b
Then, plug this and the slope into the equation
y = 1/5x - 6.4 will be the equation
Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the value of n. Round the answers to four decimal places and compare the results with the exact value definite integral.
∫9 4 √xdx,n=8.
Answer and Step-by-step explanation: The Trapezoidal and Simpson's Rules are method to approximate a definite integral.
Trapezoidal Rule evaluates the area under the curve (definition of integral) by dividing the total area into trapezoids.
The formula to calculate is given by:
[tex]\int\limits^a_b {f(x)} \, dx = \frac{b-a}{2n}[f(x_{0})+2f(x_{1})+2f(x_{2})+...+2f(x_{n-1})+f(x_{n})][/tex]
The definite integral will be:
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{9-4}{2.8}[2+2.\sqrt{5} +2.\sqrt{6} +2.\sqrt{7}+2.\sqrt{8}+3][/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{5}{16}[25.3193][/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 7.9122[/tex]
Simpson's Rule divides the area under the curve into an even interval number of subintervals, each with equal width.
The formula to calculate is:
[tex]\int\limits^a_b {f(x)} \, dx = \frac{b-a}{3n}[f(x_{0})+4f(x_{1})+2f(x_{2})+...+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})][/tex]
The definite integral will be:
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{9-4}{3.8}[2+4.\sqrt{5} +2.\sqrt{6} +4.\sqrt{7} +4\sqrt{8} +3][/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{5}{24}[40.7398][/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 8.4875[/tex]
Calculating the definite integral by using the Fundamental Theorem of Calculus:
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \int\limits^9_4 {x^{\frac{1}{2} }} \, dx[/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{2.\sqrt[]{x^{3}} }{3}[/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = \frac{2.\sqrt[]{9^{3}} }{3}-\frac{2.\sqrt[]{4^{3}} }{3}[/tex]
[tex]\int\limits^9_4 {\sqrt{x} } \, dx = 12.6667[/tex]
Comparing results, note that Simpson's Rule is closer to the exact value, i.e., gives better approximation to the exactly value calculated by the fundamental theorem.
Find the area of the region that lies inside the first curve and outside the second curve.
r= 10cos( θ)
r= 5
Answer:
Step-by-step explanation:
From the given information:
r = 10 cos( θ)
r = 5
We are to find the the area of the region that lies inside the first curve and outside the second curve.
The first thing we need to do is to determine the intersection of the points in these two curves.
To do that :
let equate the two parameters together
So;
10 cos( θ) = 5
cos( θ) = [tex]\dfrac{1}{2}[/tex]
[tex]\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}[/tex]
Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e
[tex]A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta[/tex]
[tex]A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta[/tex]
[tex]A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}[/tex]
[tex]A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}[/tex]
[tex]A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}[/tex]
[tex]A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}[/tex]
[tex]A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}[/tex]
[tex]A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}[/tex]
[tex]A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}[/tex]
The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.
411,500 science notation
Answer:
the answer is 4.115 x 10^5
Step-by-step explanation:
hope that helps
A fair die is rolled 72 times and the percentage of 6s is recorded. What is the probability that at most 10% of the rolls are 6s
Answer:
P ([tex]\hat p[/tex] ≤ 0.10)
Step-by-step explanation:
The probability in terms of statistics for this given problem can be written as follows.
Let consider X to the random variable that represents the number of 6's in 7 throws of a dice, then:
X [tex]\sim[/tex]Bin ( n = 72, p = 0.167)
E(X) = np
E(X) = 72× 0.167
E(X) = 12.024
E(X) [tex]\simeq[/tex] 12
p+q =1
q = 1 - p
q = 1 - 0.167
q = 0.833
V(X) = npq
V(X) = 72 × 0.167 × 0.833
V(X) = 10.02
V(X) [tex]\simeq[/tex] 10
∴ X [tex]\sim[/tex] N ([tex]\mu = 12, \sigma^2 =10[/tex])
⇒ [tex]\hat p = \dfrac{X}{n} \sim N ( p, \dfrac{pq}{n})[/tex]
where p = 0.167 and [tex]\dfrac{pq}{n}[/tex] = [tex]\dfrac{0.167 \times 0.833}{72}[/tex] = 0.00193
∴ P(at most 10% of rolls are 6's)
i.e
P ([tex]\hat p[/tex] ≤ 0.10)
Robert has available 400 yards of fencing and wishes to enclose a rectangular area. Express the areaAof the rectangle as a function of the widthwof the rectangle. For what value ofwis the arealargest? What is the maximum area?
Answer:
A) A = 200w - w²
B) w = 100 yards
C) Max Area = 10000 sq.yards
Step-by-step explanation:
We are told that Robert has available 400 yards of fencing.
A) we want to find the expression of the area in terms of the width "w".
Since width is "w", and perimeter is 400,if we assume that length is l, then we have;
2(l + w) = 400
Divide both sides by 2 gives;
l + w = 200
l = 200 - w
Thus, Area of rectangle can be written as;
A = w(200 - w)
A = 200w - w²
B) To find the value of w for which the area is largest, we will differentiate the expression for the area and equate to zero.
Thus;
dA/dw = 200 - 2w
Equating to zero;
200 - 2w = 0
2w = 200
w = 200/2
w = 100 yards
C) Maximum area will occur at w = 100.
Thus;
A_max = 200(100) - 100(100)
A_max = 10000 sq.yards
is 27.14159 rational or irrational
Answer:
It´s rational
Step-by-step explanation:
27,14159 = 2714159/100000
Rational
If a wind turbine makes 64 full revolutions every 1 minute, what is its angular speed?
Answer:
this is wind turbine angular speed
Step-by-step explanation:
given data
angular speed ω = 64 rpm
time = 1 min = 60 seconds
solution
we know that angular speed ω is expess as
ω = [tex]\frac{2\pi }{T}[/tex] .........................1
ω = 64 × [tex]\frac{2\pi }{T}[/tex]
ω = 6.70 rad/s
so this is wind turbine angular speed
Solve for w 98 = 7w Simplify your answer as much as possible.
Answer:
W = 14
Step-by-step explanation:
7w=98
Divide
w=98/7
Done
w=14
Hope this helps! :)
(pls mark brainliest)
Answer:
w = 14
Step-by-step explanation:
98 = 7w
98/7 = 7w/7
14 = w
Hope this helps.
what is the percentage of the total variation in candy bar sales explained by the regression model? a. 78.39% b. 88.54% c. 100% d. 48.19%
Complete question :
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
City - - - - - - - Price ($) -- - Sales
River City - - 1.30 - - - - - - 100
Hudson - - - 1.60 - - - - - 90
Ellsworth - - - 1.80 - - - - - 90
Prescott - - - - 2.00 - - - - 40
Rock Elm - - 2.40 - - 38
Stillwater - - 2.90 - - 32
Answer:
78.39%
Step-by-step explanation:
Given the data :
Price (X) :
1.30
1.60
1.80
2.00
2.40
2.90
Sales (y) :
100
90
90
40
38
32
The percentage of the total variation in candy bar sales explained by the regression model can be obtained from the value of the Coefficient of determination(R^2) of the regression model. The Coefficient of determination is a value which ranges between 0 - 1 and gives the proportion of variation in the dependent variable which can be explained by the dependent variable.
R^2 value is obtained by getting the squared value of R(correlation Coefficient).
The R value obtained using the online R value calculator on the data is : - 0.8854
Hence, R^2 = (-0.8854)^2 = 0.7839
Expressing 0.7839 as a percentage ;
0.7839 × 100 = 78.39%
evaluate the following -3 - (-8)
Answer:
The answer to the problem is 5.
Answer:
5Step-by-step explanation:
[tex]-3-\left(-8\right)\\\\\mathrm{Apply\:rule}\:-\left(-a\right)=a\\=-3+8\\\\\mathrm{Add/Subtract\:the\:numbers:}\:\\-3+8\\=5[/tex]
Find the slope and y-intercept (if possible) of the equation of the line. 14x − 6y = 66
Answer:
The slope would be 7/3
The y-intercept would be -11
Step-by-step explanation:
To find the answer rearrange your equation.
First subtract 14x from both sides, getting -6y= -14x + 66.
Then divide both sides by -6, getting y= 7/3x -11
This is now written in y=mx+b form. Your m is your slope and b is the y intercept!
which equation represents a line that is parallel to y= -4x+3 and passes through the point (-3,2)
Answer:
The answer is
[tex]y = - 4x - 14[/tex]Step-by-step explanation:
Equation of a line is y = mx + c
where
m is the slope
c is the y intercept
To find the equation of the parallel line we must first find the slope of the original line
From the question
y = - 4x + 3
Comparing with the general equation above
Slope / m = - 4
Since the lines are parallel their slope are also the same
Slope of parallel line = (-3 , 2) and slope
- 4 is
[tex]y - 2 = - 4(x + 3) \\ y + 2 = - 4x - 12 \\ y = - 4x - 12 - 2[/tex]We have the final answer as
[tex]y = - 4x - 14[/tex]Hope this helps you
Answer:
y=-4x-14
Step-by-step explanation:
Simplify and Show all of your work please!
Answer:
[tex]\Huge \boxed{\mathrm{9}}[/tex]
Step-by-step explanation:
[tex]\Rightarrow \displaystyle \frac{18 \div 2 * 3}{5-2}[/tex]
Dividing first.
[tex]\Rightarrow \displaystyle \frac{9 * 3}{5-2}[/tex]
Multiplying and subtracting.
[tex]\Rightarrow \displaystyle \frac{27}{3}[/tex]
Division.
[tex]\Rightarrow 9[/tex]
Find the equation of the line using the point-slope formula. Passes through the point (−5, 8) and is parallel to the graph y = 4/5x+1
Answer: y=4/5x+12
Step-by-step explanation:
Slope-intercept form: y=ax+b, where a is the slope and b is the y-intercept
When a line is parallel to another line, they will have the same slope. So the slope of the new line is also 4/5.
y=4/5x+b
To find the y-intercept, we simply plug in the point it passes through.
y=4/5x+b
8=4/5(-5)+b
8=-4+b
b=12
y=4/5x+12
Hope this helps!! :)
Please let me know if you have any question
Erin is considering joining one of 2 clubs. Club A has no registration fee, but charges $105 per month. Club B charges members $80 per month plus a one-time registration fee of $375. For how many months is club A the cheaper option? Use system of equations
Answer:
4 months
Step-by-step explanation:
Club A
registration fee: $0
monthly fee: $105
After every month, the total cost increases by $105.
month 0: $0
month 1: $105
month 2: $210
month 3: $315
month 4: $420
month 5: $525
month 6: $630
Club B
registration fee: $375
monthly fee: $80
Notice how Club B's total reaches Club A's total after 2 months.
month 0: $375
month 1: $455
month 2: $535
month 3: $615
month 4: $695
n ΔABC, AB = 10 and BC = 5. Which expression is always true? A. 5 < AC < 10 B. AC = 5 C. 5 < AC < 15 D. AC = 10
Answer:
A. 5 < AC < 10
Step-by-step explanation:
If ∆ABC is a right angled triangle, we use the Pythagoras formula:
c² = a² + b²
Where c = longest side
When given sides AB, AC and BC, the formula becomes:
AB² = AC² + BC²
Where AB = Longest side
In the question,
AB = 10 and BC = 5.
10² = AC² + 5²
AC² = 10² - 5²
AC² = 100 - 25
AC² = 75
AC = √75
AC = 8.6602540378
Therefore, the expression that is always true = A. 5 < AC < 10
Round to the nearest cent.
6. $10.407
Answer:
the answer is 10.41. If you have a number 5 or more you round the nearest number on the left up 1 if it's 4 or less it stays the same it doesn't go up or down.
algebra 2
50 POINTS
HELP
Answer:
{-3, 2}U{2, 5}
Step-by-step explanation:
For an equation to be negative, it would need to be in a negative range (below the x-axis or the coordinates are negative y-values). Therefore, we can examine this question and see that the graph is negative when the function crosses the x-axis at -3 and it remains negative until you reach 2 on the x-axis.
Therefore, the first set of negative values is (-3, 2).
Secondly, applying the same logic as before, the function decreases at 2 and then touches the x-axis again at 5. Therefore, the second negative value would be (2, 5).
The negative values are {-3, 2}U{2, 5}.
Answer:
{-3, 2}U{2, 5}
Step-by-step explanation:
I really help worth these question.
Answer:
[tex] \frac{7}{3} [/tex]
Step-by-step explanation:
Given that,
p = -6,
q = 6
r = -19
Plug in the above values to evaluate the expression, [tex] \frac{\frac{q}{2} - \frac{r}{3}}{\frac{3p}{6} + \frac{q}{6}} [/tex]
[tex] \frac{\frac{6}{2} - \frac{(-19)}{3}}{\frac{3(-6)}{6} + \frac{6}{6}} [/tex]
[tex] \frac{\frac{3}{1} - \frac{(-19)}{3}}{\frac{-3}{1} + \frac{1}{1}} [/tex]
[tex] \frac{\frac{9 -(-19)}{3}}{3 + 1} [/tex]
[tex] \frac{\frac{28}{3}}{4} [/tex]
[tex] \frac{28}{3}*\frac{1}{4} [/tex]
[tex] \frac{28*1}{3*4} [/tex]
[tex] \frac{7*1}{3*1} [/tex]
[tex] \frac{7}{3} [/tex]
What will be the remainder when 6x ^5+ 4x^4 -27x^
3
- 7x² + 27x + 3/2 is divided by (2x^2 - 3)
^2
Answer:
Remainder = (3145/8)x - 408
Step-by-step explanation:
We want to find the remainder when 6x^(5) + 4x⁴ - 27x³ - 7x² + 27x + 3/2 is divided by (2x² - 3)²
Let's expand (2x² - 3)² to give ;
(2x - 3)(2x - 3) = 4x² - 6x - 6x + 9 = 4x² - 12x + 9
So,we can divide now;
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
First of all, we'll divide the term with the highest power inside the long division symbol by the term with the highest power outside the division symbol. This will give;
3/2x³
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
We now subtract the new multiplied term beneath the original one from the original one to get;
3/2x³
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³+27x +3/2
We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;
(3/2)x³ + (11/2)x²
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³+27x +3/2
22x⁴-66x³ + (99/2)x²
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
We now subtract the new multiplied term beneath the immediate one from the immediate one to get;
(3/2)x³ + (11/2)x²
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³-7x²+27x +3/2
22x⁴-66x³ + (99/2)x²
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
(159/2)x³-(113/2)x²+27x+(3/2)
We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;
(3/2)x³ + (11/2)x² + (159/8)x
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³-7x²+27x +3/2
22x⁴-66x³ + (99/2)x²
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
(159/2)x³-(113/2)x²+27x+(3/2)
(159/2)x³-(477/2)x²+(1431/8)x
We now subtract the new multiplied term beneath the immediate one from the immediate one to get;
(3/2)x³ + (11/2)x² + (159/8)x
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³-7x²+27x +3/2
22x⁴-66x³ + (99/2)x²
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
(159/2)x³-(113/2)x²+27x+(3/2)
(159/2)x³-(477/2)x²+(1431/8)x
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
182x²-(1215/8)x + (3/2)
We'll now divide the term in new polynomial gotten with the highest power by the term with the highest power outside the division symbol. This gives;
(3/2)x³+(11/2)x²+(159/8)x+(91/2)
______________________
4x²-12x+9 |6x^(5)+4x⁴-27x³-7x²+27x+3/2
6x^(5)-18x⁴-(27/2)x³
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
22x⁴+(27/2)x³-7x²+27x +3/2
22x⁴-66x³ + (99/2)x²
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
(159/2)x³-(113/2)x²+27x+(3/2)
(159/2)x³-(477/2)x²+(1431/8)x
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
182x²-(1215/8)x + (3/2)
182x²-545x + 819/2
We now subtract the new multiplied term beneath the immediate one from the immediate one to get;
182x² - (1215/8)x + (3/2) - 182x² + 545x - 819/2 = (3145/8)x - 408
Remainder = (3145/8)x - 408
PLEASE ANSWER I REALLY NEED AND PEOPLE DONT ANSWER :(
1: identify the terms, coefficients, and constants in the expression 14x + 19.
Answer:
term=14x+19
cofficient=14+19
constant=19
Answer:
Terms :
14x , 19
Coefficients:
14
Constants:
19
Hope it helps.
Tony ran 1/2 of a mile for 1/4 of an hour. How many miles per hour did he run? A)1.0 B)2.0 C)3.0 D)4.0 E)5.0
Answer:
2.0
Step-by-step explanation:
Answer:
B. 2.0 miles
Step-by-step explanation:
Tony ran 1/2 of a mile for 1/4 of an hour.
First, to make it easier, change each fraction into decimals:
1/2 = 0.5
1/4 = 0.25
It takes Tony 0.25 hours to run 0.5 miles.
You are solving for 1 hours worth. Multiply 4 to both terms:
0.25 hr x 4 = 1 hr
0.5 miles x 4 = 2.0 miles
B. 2.0 miles is your answer.
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The equation a = 640 s gives the relationship between s square miles and a acres. Pam owns 4.5 square miles of farmland. How many acres does she own? a. 2,880 acres b. 288 acres c. 0.7 acres d. 7.03 acres
Answer:
A. 2880 acres
Step-by-step explanation:
Formula: a = 640 s
Given information: s = 4.5
--> a = 640 x 4.5 = 2880 (acres)
Which x values is the graph below discontinuous
Answer:
-3,-1,1,3,5
Step-by-step explanation:
Answer I need help !!!!!!!!!!!
Answer:
Pay for the day = $ 123.25
Step-by-step explanation:
From the question given:
Monday morning:
Time in: 8:15
Time out: 12:15 pm
Monday afternoon:
Time in: 13:00
Time out: 17:30
Pay = $ 14.5 /hr
Next, we shall determine the number of hours of work in the morning. This is illustrated below:
Time in: 8:15
Time out: 12:15 pm
Difference in time = 12:15 – 8:15 = 4 hrs
Next, we shall determine the pay for the work done in the morning. This can be obtained as follow:
Pay = $ 14.5 /hr
Pay for work done in the morning
= 4 × 14.5 = $ 58
Next, we shall determine the number of hours of work in the afternoon. This is illustrated below:
Time in: 13:00
Time out: 17:30 pm
Difference in time = 17:30 – 13:00 = 4 hrs 30 minutes
Next, we shall convert 4 hrs 30 minutes to hours. This is illustrated below:
60 minutes = 1 hr
30 minutes = 30/60 = 0.5 hrs.
Therefore,
4 hrs 30 minutes = 4 + 0.5 = 4.5 hrs
Next, we shall determine the pay for the work done in the afternoon. This can be obtained as follow:
Pay = $ 14.5 /hr
Pay for work done in the afternoon
= 4.5 × 14.5 = $ 65.25
Finally, we shall determine the pay for the day as follow:
Pay for work done in the morning
= $ 58
Pay for work done in the afternoon
= $ 65.25
Pay for the day = pay for morning + pay for afternoon
Pay for the day = $ 58 + $ 65.25
Pay for the day = $ 123.25
Therefore, the pay for the day is
$ 123.25
The electrical resistance of a wire, R, varies directly as its length, L, and inversely as its cross sectional area, A. If the resistance of a wire is 0.08 ohm when the length is 100 ft and its cross-sectional area is 0.05 in^2 , what is the resistance of a wire whose length is 4000 ft with a cross-sectional area of 0.02in^2 ? a) Write the variation equation. (R=k(l/a) b) Determine the value of the quantity indicated.
Answer:
see below
Step-by-step explanation:
a R=k(L/A)
Substitute what we know into the equation
.08 = k (100/.05)
.08 = k 2000
Divide each side by 2000
.08/2000 = k
.00004 = k
R=.00004 (L/A)
b R=.00004 (L/A)
We know L = 4000 and A = .02
R=.00004 (4000/.02)
R = 8