Please find answers of 14 a, b, c and d​

Answer:

(-4,0) and (4,0)

Step-by-step explanation:

The x-intercepts occur for values of x where f(x) = 0.

f(x) = 0 when the nominator ([tex]x^{2} - 16[/tex]) = 0.

[tex]x^{2} -16[/tex] is a difference of squares,
so you can factor it as (x + 4)(x – 4)

When x = –4 or x = 4, the nominator is 0, and f(x) = 0

[tex]a(0) = 1 \: \: \: \: \: \: b(0) = 2 \: \: \: \: \: c(0) = 3 \\ a(1) = b(0) + c(0) = 2 + 3 = 5 \\ b(1) = a(0) + c(0) = 1 + 3 = 4 \\ c(1) = a(0) + b(0) = 1 + 2 = 3 \\ \\ a(2) = b(1) + c(1) = 4 + 3 = 7 \\ b(2) = a(1) + c(1) = 5 + 3 = 8 \\ c(2) = a(1) + b(1) = 5 + 4 = 9[/tex]

[tex]a(3) = b(2) + c(2) = 8 + 9 = 17\\ b(3) = a(2) + c(2) =7 + 9 = 16 \\ c(3) = a(2) + b(2) = 7 + 8 = 15 \\ \\ a(4) = b(3) + c(3) = 16 + 15 = 31 \\ b(4) = a(3) + c(3) = 17 + 15 = 32 \\ c(4) = a(3) + b(3) = 17 + 16 = 33[/tex]

[tex]a(5) = 32 + 33 = 65 \\ b(5) = 31 + 33 = 64 \\ c(5) =31 + 32 = 63 \\ \\ a(6) = 64 + 63 = 127 \\ b(6) = 65 + 63 = 128 \\ c(6) = 65 + 64 = 129 \\ [/tex]

[tex]a(7) = 128 + 129 = 257 \\ b(7) = 127 + 129 = 256 \\c (7) = 127 + 128 = 255 \\ \\ a(8) = 256 + 255 = 511 \\ b(8) = 257 + 255 = 512 \\ c(8) = 257 + 256 = 513[/tex]

[tex]a(9) = 512 + 513 = 1025 \\ b(9) = 511 + 513 = 1024 \\ c(9) = 511 + 512 = 1023 \\ \\ a(10) = 1024 + 1023 = 2047 \\ b(10) = 1025 + 1023 = 2048 \\ c(10) = 1025 + 1024 = 2049[/tex]

[tex]a(n) + b(n) + c(n) = \\ 2(a(n - 1) + b(n - 1) + c(n - 1)) \\ 6 \times 2 {}^{n } [/tex]

[tex]6 \times 2 {}^{n} > 100 \: 000 \\ 2 {}^{n} > \frac{100 \: 000}{6} \\ n > log {}^{2} ( \frac{100 \: 000}{6} ) \\ n > 14.02468 \\ n = 15[/tex]

The area of the minor sector DFE is gotten as; D: 66.1 square cm

Formula for area of sector is;

A = (θ/360) * πr²

We are given;

minor angle; θ = 100°

radius; r = 8.7 cm

Thus;

A =  (100/360) * π(8.7²)

A = 66.1 square cm

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The decision would be to fail to reject the null hypothesis and conclude that there is insufficient evidence.

The formula is fx/n

To get fx we have to sum all of the values of x = 2430161

The total number in the data set = 35

Mean = 2430161/35

= 69433.17

This test is a left tail test.

sd = 8500

H0: u = 70000

H1 : U < 70000

z0.05 = 1.645

The decision would be to reject if z < critical value

Z= (69433.17 - 70000)/(8500/√35)

= -0.395

P-value can be gotten from the z score = 0.348

Given that the p value is > 0.05, we would have to fail to reject the null. The conclusion would be that there is insufficient evidence to support claim.

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The end behavior of the Polynomial f(x) = -4x⁶ + 6x² - 52 is;

B. f(x) is an even function so both ends of the graph go in the same

direction

The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity.

Now, we are given the polynomial;

f(x) = -4x⁶ + 6x² - 52

Let us first test if the polynomial is even;

f(1) = -4(1)⁶ + 6(1)² - 52

f(1) = -50

Similarly;

f(-1) = -4(-1)⁶ + 6(-1)² - 52

f(-1) = -50

Since f(1) = f(-1), we can say that both ends of the graph go in opposite

directions.

Lastly, the leading coefficient is negative because it is -4 and as such it means that the left end goes down.

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We would reach our office by 7:34 a.m.

The average speed is given as 30 miles per hour.

Let us convert this into miles per minute.

30 miles/1 hour = 30 miles/60 minutes

= 0.5 miles per minute

The distance between the house and the office is 12 miles.

Therefore, we can find what it would take to reach the office by dividing the distance by the average speed:

Time taken = distance/average speed

= 12 miles/0.5 miles per minute

= 24 minutes

The time is given to be 7:10 a.m. Therefore we would reach our office by 7:34 a.m.

Therefore, we would reach our office by 7:34 a.m.

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Answer:

5 units

Step-by-step explanation:

Given the following question:

To find the answer to this question we will use the formula to calculate distance.

(6, 5) = (x1, y1)
(-3, 1) = (x2, y2)

[tex]d=\sqrt{(x1-x2)^2+(y1-y2)^2}[/tex]
[tex]d=\sqrt{(-6-(-3))^2+(5-1)^2}[/tex]
[tex]-6+3=-3[/tex]
[tex]5-1=4[/tex]
[tex]d=\sqrt{(-3^2+4^2)}[/tex]
[tex]-3^2=-3\times-3=9[/tex]
[tex]4^4=4\times4=16[/tex]
[tex]d=\sqrt{9+16}[/tex]
[tex]9+16=25[/tex]
[tex]\sqrt{25}[/tex]
[tex]\sqrt{25} =5\times5=25[/tex]
[tex]=5[/tex]

Which means the distance between the two points is "5 units."

Hope this helps.

[tex]\huge\underline{\red{A}\green{n}\blue{s}\purple{w}\pink{e}\orange{r} →}[/tex]

Step-by-step explanation:

5x:9 = 7:3

→ 5x/9 = 7/3

→ x = (7 × 9)/(3 × 5)

→ x = 21/5

[tex]\sqrt{3-2\sqrt{2} }[/tex] = 0.06

(2x)ˣ = 25√5

[tex]\sqrt{3-2\sqrt{2} }[/tex]  = [tex]\sqrt{3}-\sqrt{2\sqrt{2} }[/tex] = 1.73205080757 - 1.67428223427 = 0.06

4ˣ . 4ˣ⁻¹ = 24

4ˣ . 4ˣ / 4 = 24

4ˣ(1 - 1 / 4) = 24

4ˣ(3 / 4) = 24

multiply both sides by 4 / 3

4ˣ = 32

2²ˣ = 2⁵

2x = 5

x = 5 / 2 = 2.5

Therefore,

(2x)ˣ = 25√5

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If the time taken by 14 inches wide is one hour then the blade of 12 inches wide will take 1 hour 17 minutes to mow a lawn.

What is width?

The width of something is the measurement of one side of a body. It is generally less than length of that body or shape.

How to calculate time?

We have been given time of 1 hour if blade is 14 inches wide and we have to calculate time taken by 12 inches wide blade:

It will be as 14/12* 1 hour

=14/12

=1.167

=1.17 hour

Hence the time taken by the blade of 12 inches wide will be 1 hour and 17 minutes.

Answer:

70 mins

Step-by-step explanation:

This is called inverse proportion.

1hr : 14in

x hr : 12in

Ask yourself, what do you do to 14 to get 12? To find this out, we do:

14 / 12 = 1.16666667

Since the blade is smaller, it will take longer. So, we times 1hr by 1.16666667 to get

1.16666667 hrs =

1.17 hours (rounded)

This is 70.2 mins, or 1hr 10 mins

Hope this helps!

Answer:

[tex]x^{3} ( x + 1 ) + y^{3} (x+1)\\= (x+1)(x^{3} + y^{3})[/tex]

Answer:

Step-by-step explanation:

YOU PLUS ALL OF THEM

Answer:

B. (3x - 9) (x + 3i)

Step-by-step explanation:

[tex]\bf 3 {x}^{2} = + 27[/tex]

Therefore, B is your answer!!!!

Answer:

1:3.

Step-by-step explanation:

Probability(getting > 3 on one roll) = 3/6 = 1/2.

So on 2 cubes rolled the probability   of getting > 3 is 1/2 * 1/2 = 1/4.

As a ratio this is 1:3.

abc =32 ft³  is the maximum volume of an open rectangular box (with no top face) if its surface area is 1 square foot.

A rectangle with all equal sides, commonly known as a square.  Multiplying the length by the length is the. Using L as the length of each side, solve for L X L = L2,

The maximum volume of an open rectangular box (with no top face) if its surface area is 1 square foot This Lagrange multiplier optimization is standard. If the box has a base of a, a height of c, and an area constraint of ab+2ac+2bc−48=0  we wish to optimize V= abc.

L(a,b,c,λ)= abc−λ(ab+2ac+2bc−48)

The four partial derivatives are zero at an ideal position, so:

δLδa=bc−λ(b+2c)=0

δLδb=ac−λ(a+2c)=0

δLδc=ab−λ(2a+2b)=0

Plus the restriction. The first two enlighten:

λ=bcb+2c=aca+2c

Consequently,  b(a+2c)=a(b+2c)  implies to b=a. The third partial, where b=a, now informs us that   a2=4aλ  and so  λ=a/4  nd by using this information in the second partial, we obtain  4c=a+2c  which informs us that c=a/2 .

Now that we've inserted these b and c expressions into the constraint, we get [tex]3a^2=1[/tex]  which means that a=4 feet, b=4 feet, and c=2 feet.

The maximum volume is therefore, abc=32 ft³

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The value of integers a and b is 1 and 2 and the value of a/b is 1/2.

According to the statement

we have  a given a expression 4k+8 / 4 and we have to simplify this expression in the form of ak+b and find the value of a/b where a and b are the integers.

An integer is a whole number (not a fractional number) that can be positive, negative, or zero.

Here 4k+8/4 becomes

4k/4 +8/4  -(1)

and general form is

ak+b  -(2)

Now, compare these equations then we get

a = 1 and b = 2.

Now we find the value of a/b

then

a/b = 1/2.

So, The value of integers a and b is 1 and 2 and the value of a/b is 1/2.

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The surface area of the composite figure is 524 square meters

The given composite figure is made up of a triangular prism and a rectangular prism. The surface area of the figure is given as:

Surface area = 2(16*5+5*4+16*4) + (16*6) + 2(50)

Take the sum

Surface area = 2(80+20+64) + 96 + 100

Surface area = 2(164) +196

Surface area = 328 + 196

Surface area = 524 square meters

Hence the surface area of the composite figure is 524 square meters

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A)  Rotate by 90° counterclockwise.

B) Reflection transformation about the line y = 5.

C) Parallel Lines are C₂D₂ and A₂B₂; EF and E₁F₁.

D) Perpendicular lines are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Line segments with slope of 0 are; AB, C₂D₂, A₂B₂, EF and E₁F₁

F) Line segments on the boat that have an undefined slope are; Lines A₁B₁, DF and DC.

G) Slope of Line ED = 1

A) The blue segment CD is seen on the graph as a perpendicular line with 2 units while the line segment C₂D₂ is seen as a horizontal line. Thus, to match CD with C₂D₂, we will rotate by 90° counterclockwise.

B) The transformations that would be used on the blue triangle to get it to match with the red triangle is a reflection transformation about the line y = 5.

C) The line segments that are parallel to each other are; C₂D₂ and A₂B₂; EF and E₁F₁.

D) The line segments that are perpendicular are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Horizontal lines that are parallel to the x-axis have zero slope. Thus, AB, C₂D₂, A₂B₂, EF and E₁F₁ all have zero slopes.

F) Undefined slope is the slope of a vertical line. Thus, Lines A₁B₁, DF and DC have undefined slopes.

G) Slope of ED = (5 - 4)/(-10 - (-11))

Slope of ED = 1/1

Slope = 1

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Question 2

1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)

Completing the squares, we have that:

The equation of a quadratic function, of vertex (h,k), is given by:

y = a(x - h)² + k

In which a is the leading coefficient.

For the parabola of standard form y = x² - 6x + 17, we have that it can be written as follows, completing the squares:

y = x² - 6x + 17 = (x - 3)² + 8, hence the extreme value is (3,8).

For the parabola of standard form y = x² + 8x + 21, we have that it can be written as follows, completing the squares:

y = x² + 8x + 21 = (x + 4)² + 5, hence the extreme value is (-4,5).

For the parabola of standard form y = x² - 16x + 60, we have that it can be written as follows, completing the squares:

y = x² - 16x + 60 = (x - 8)² - 4, hence the extreme value is (8,-4).

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The interface validation that would help to confirm that you meet the minimum age is the range check.

This is the term that is used to refer to the check that is done in such a way that it shows the data that are in a particular range.

This would help to ensure that there is a specified upper and lower boundary for the values to fall into.

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which of the following interface validation checks is useful in verifying that the birth data you entered on your job application indicates that you meet the minimum age requirement to serve alcohol?

Consistency check

Format check

Completeness check

Range check

The volume of the cylinder is 301.593 cubic centimeter.

Volume exists as a scalar quantity representing the amount of three-dimensional space enclosed by a sealed surface. For example, the space that a substance or 3D shape inhabits or contains. The volume stands often quantified numerically utilizing the SI-derived unit, the cubic meter.

A cylinder contains traditionally been a three-dimensional solid, one of the most essential curvilinear geometric shapes. Geometrically, it has been regarded as a prism with a circle as its base.

The dimensions of the cylinder is given as - 8 cm and 12 cm .

The formula of the volume of the cylinder is given as -

V = π * (8) * (12)

Thus the volume is = 96π ≈ 301.593 cubic centimeter.

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Answer:

Obtuse, scalene.

Step-by-step explanation:

In ΔSVU, one angle is 120° which is obtuse angle.

So, ΔSVU is a obtuse triangle.

All the sides  have different measurements. So, It is a Scalene triangle.

He is  2√19 miles far way from starting point.

Suppose Billy starts at point A, turns at point B, and ends at point D, as shown below.

If Billy turns 60◦ northward and walks six miles, then we can draw a 30 − 60 − 90 triangle whose hypotenuse is 6 miles

It follows that Billy traveled 6/2 = 3 miles eastward during these 6 miles, and that he traveled 3√3 miles northward during these 6 miles. In total, Billy traveled 4 + 3 = 7 miles  eastward and 3√ 3 miles northward.

By the Pythagorean Theorem, the distance from his starting point is q:

(7)2 + (3√3)2

=√49 + 27

= √76

= 2√19 .

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Answer:

[tex]log_{2}[/tex] 20

Step-by-step explanation:

using the rule of logarithms

loga + logb = logab

then

[tex]log_{2}[/tex]4 + [tex]log_{2}[/tex]5

= [tex]log_{2}[/tex] (4 × 5)

= [tex]log_{2}[/tex]20

Answer:

Below.

Step-by-step explanation:

x^2 + 4x  = 18

x^2 + 4x + (2)^2 = 15 + 2^2

x^2 + 4x + 4 = 18 + 4

(x + 2)^2 = 22

x  + 2 = +/- √22

x = -2 + √22 and x =  -2 + √22

x = 2.7  and -6.7  to nearest tenth.

Answer:

- 226

Step-by-step explanation:

ARITHMETIC SEQUENCE.

Number of term of an Arithmetic progressions has the formular.

Tn = a + ( n - 1 ) d

From the question,

First term ( a ) = 70

common difference = T2 - T1 = 63 - 70 = -7

For the 49th term

T49 = a + 48d

= 70 + 48 ( -7 )

= 70 - 336 = - 226

Using linear functions, it is found that over a range of x < 10000 she will make more from the Comfort over Style job than she will from the Heels for Fashion Slaves job

A linear function is modeled by:

y = mx + b

In which:

Considering the fixed salary as the intercept and the earning per shoe sold the slope, the functions for her salaries at each situation is:

He will make more from the Comfort over Style job than she will from the Heels for Fashion Slaves job when:

C(x) > H(x)

52000 + 3.25x > 37000 + 4.75x

1.5x < 15000

x < 15000/1.5

x < 10000

The range is x < 10000.

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If [tex]x[/tex] and [tex]y-1[/tex] have the same sign, then either

[tex]x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9[/tex]

or

[tex]x<0,y<1 \implies 2|x| + 3|y-1| = -2x - 3(y-1) = 6 \implies 2x + 3y = -3[/tex]

If [tex]x[/tex] and [tex]y-1[/tex] have opposite sign, then

[tex]x>0,y<1 \implies 2|x| + 3|y-1| = 2x - 3(y-1) = 6 \implies 2x -3y = 3[/tex]

or

[tex]x<0,y>1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9[/tex]

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

[tex]R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}[/tex]

The area of [tex]R[/tex] is given by the double integral

[tex]\displaystyle \iint_R dx\,dy[/tex]

To compute the area, change the variables to

[tex]\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}[/tex]

The Jacobian for this transformation is

[tex]J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}[/tex]

with determinant [tex]\det(J) = -\frac1{12}[/tex]. Then the integral transforms to

[tex]\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du[/tex]

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

[tex]\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}[/tex].

The length of the missing length in the triangle is 53.6 units

The given triangle is a right triangle.

Right triangles are triangles that have a measure of angle that is 90 degrees.

The missing side is x, and the value of x is calculated using the following sine rule

We have

sin(angle) = opposite/hypotenuse

This gives

sin(66) = 49/x

Divide both sides by sin(66)

1 = 49/x sin(66)

Multiply both sides by x

x = 49/sin(66)

Divide the expression

x = 53.6

Hence, the length of the missing length is 53.6 units

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Answer:

  18 cans

Step-by-step explanation:

The number of cans of paint will be the least integer greater than or equal to the number required to cover the area of the trapezoid.

The area of the wall is given by the formula ...

  A = 1/2(b1 +b2)h

where b1 and b2 are the bases, and h is the height.

We are given the values b1=20 ft, b2=24 ft, h=16 ft, so the area is found to be ...

  A = 1/2(20 ft +24 ft)(16 ft) = 352 ft² . . . . . area of the wall

Each can of paint covers 20 ft², so n cans of paint will cover 20n square feet. We want to find the minimum integer value of n that satisfies ...

  20n ≥ 352

  n ≥ 17.6 . . . . . . . . . divide by 20

The next larger integer is 18.

The painter will need to buy 18 cans of paint.