If 111 is divided into pieces that are each \dfrac19
9
1
​
start fraction, 1, divided by, 9, end fraction of a whole, how many pieces are there?
1\div\dfrac19=1÷
9
1
​
=1, divided by, start fraction, 1, divided by, 9, end fraction, equals

The equation p=4+3y is maximised at (4,5) and the maximised value is 31.

Given equation p=4x+3y and constraints x>0,x<4, y<5.

We have to maximise the equation p=4x+3y with constraints x>0,x<4,y<5.

To find out the points we have to make graphs of the constraints first.

From the graph we have found that points are (0,0),(0,5),(4,0),(4,5).

The above points are common points of graphs of all constraints.

The value of equations are as under:

At (0,0) , p=4*0+3*0=0

At (0,5), p=4*0+3*5=15

At (4,0),p=4*4+3*0=16

At (4,5),p=4*4+3*5=16+15=31

Maximum value of p is 31 and that is for (4,5).

Hence the equation is maximised at (4,5) and the maximum value is 31.

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The minimum value of z is -12.

Given that the system of linear programming

z=-3x+5y

x+y≥-22

x-y≥-4

x-y≤2

Firstly, we will convert the given constraints into equality as

x+y=-22

x-y=-4

x-y=2

Now, we will take the equation x+y=-22 and taking x as 0 as find y and taking y as 0 to find x, we get

If x=0 then y is

0+y=-22

y=-22

If y=0 then x=-22

So, the points lie on the line x+y=-22 is (0,-22) and (-22,0).

Further, we will take the equation x-y=-4 and taking x as 0 as find y and taking y as 0 to find x, we get

If x=0 then y=4

If y=0 then x=-4

So, the points lie on the line x-y=-4 is (0,4) and (-4,0).

Furthermore, we will take the equation x-y=2 and taking x as 0 as find y and taking y as 0 to find x, we get

If x=0 then y=-2

If y=0 then x=2

So, the points lie on the line x-y=2 is (0,-2) and (2,0).

now, we will find the feasible region of the given constraints as shown in figure.

We will find that each constraint will contain origin or not by substituting (0,0), we get

0+0≥-22 containing origin

0-0≥-4 containing origin

0-0≤2 containing origin

From the graph we can see that the minimum value of x is -13 and minimum value of y is -12.

z=-3x+5y

z=-3x+3y+2y

z=-3(x-y)+2y

minmum value of x-y from the given constraints is

-4≤x-y≤2

From here minimum value of x-y is -4

So, z=-3(-4)+2y

z=12+2y

For the minimum value of z the value of y should be minimum.

From the graph we can see the minimum value of y=-12

So, z=12+2(-12)

z=12-24

z=-12

Hence, the minimum value of z for the system of Linear Programming:

z=-3x + 5y, x+y≥-22, x-y≥-4, x - y ≤ 2 is -12.

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The missing angle of the right trapezoid is 126°.

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. The trapezoid is a quadrilateral that presents at least one set of parallel sides and the sum of the internal angles is 360°. The trapezoid classifications are:

From the figure, it is possible to see that the image has two right angles. Thus, you have a right trapezoid.

Therefore, from the sum of internal angles, you can find the missing angle:

u+54+90+90=360

u=360-234

u=126

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The correct answer is (40, 35) because by plugging in,  40 +35 = 75 and

(40 x 5) + (35 x 3) > 300

Word problem can be solved by interpreting the problem into its best fit equation. The equation may be linear, quadratic or simultaneous equations.

Given that

5x + 3y > 300

x + y = 75

Let us first assume that 5x + 3y = 300. That is,

5x + 3y = 300

x + y = 75

Eliminate y by multiplying equation 2 by 3

5x + 3y = 300

3x + 3y = 225

2x = 75

x = 75/2

x = 37.5

Substitute x in equation 2

37.5 + y = 75

y = 75 - 37.5

y = 37.5

The correct answer is (40, 35) because 40 +35 = 75 and

(40 x 5) + (35 x 3) > 300

Therefore,  the solution to the system is (40, 35)

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Answer:

correct answer is 110x + 29y = 333

Step-by-step explanation:

The sum of areas of circle C and circle D exists 98 square units.

Given: Radius of circle = 7 units

Area of circle [tex]=\pi r^2[/tex] ..............(1)

Where, r be the radius of circle

The radius of circle C, r = 7 units

(1) becomes

Area of circle, C [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Area of circle, D [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Sum of areas of circles C and D,

= Area of circle C + Area of circle D

= 49 + 49

= 98 units²

The sum of areas of circles C and D exist 98 units².

Therefore, the correct answer is option d) 98 units².

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Step-by-step explanation:

we first divide fourteen by four

14/4= 3.5

then we change 0.5 to hours by multiplying by 60

0.5 * 60 = 30

so in conclusion it would take 3hrs 30mins to finish each activity

Question 4

[tex]\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}[/tex]

Question 5

[tex]\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}[/tex]

Answer:

no

Step-by-step explanation:

using the converse of Pythagoras' identity

if the square on the longest side is equal to the sum of the squares on the other 2 sides then the triangle is right.

longest side = 11 and 11² = 121

7² + 9² = 49 + 81 = 130

since 121 ≠ 130 then sides do not form a right triangle

None of the expressions is an identity

The expressions are given as:

I. y=x

II. 4 = 3x^2 + 2

III. x=x

In algebra, there are three identities; and they are

None of the given expression take the above forms

Hence, none of the expressions is an identity

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The cost of kilograms of almonds is $8.5 if 4 kilograms of almonds cost $34.

According to the questions,

4 kilograms of almonds cost $34. In order to find one kilograms of almonds  only if we divide 34 by 4.

= [tex]\frac{34}{4}[/tex] = 8.5

Hence, the cost of kilograms of almonds is $8.5 if 4 kilograms of almonds cost $34.

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The domain of the function is [1,  ∞) and the range is [0,  ∞)

The equation of the function is given as:

g(x) = 3√x - 1

The domain is the possible values of x.

Set the radicand to greater than or equal to 0

x - 1 >= 0

Add 1 to both sides

x >= 1

This can be rewritten as:

[1,  ∞)

Breakdown

g(x) = 3√x - 1

x - 1 >= 0

x >= 1

[1,  ∞)

The range is the possible values of y.

Set the radicand to 0

g(x) = 3 * 0

Evaluate the product

g(x) = 0

This can be rewritten as:

[0,  ∞)

Breakdown

g(x) = 3√x - 1

g(x) = 3 * 0

g(x) = 0

[0,  ∞)

Hence, the domain of the function is [1,  ∞) and the range is [0,  ∞)

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Answer:

Correct answer is D: [1, ∞) and R: (-∞, 0]

Step-by-step explanation:

Insert the given function on desmos to observe the following explanation.

The domain (x values) begin at 1, and go on to positive infinity (the squared bracket indicates a stop/start at that certain point). The range goes into negative infinity (y values) and begins at the y value of 0, so it does not go above that point on the y axis (hence the squared bracket).

Point-slope form

[tex]y-2=4(x+1)[/tex]

Slope-intercept form

[tex]y-2=4x+4 \\ \\ y=4x+6[/tex]

Standard form

[tex]4x-y+6=0[/tex]

Answer:

[tex]\textsf{Point-slope form}: \quad \sf y-2=4(x+1)[/tex]

[tex]\textsf{Slope-intercept form}: \quad \sf y=4x+6[/tex]

[tex]\textsf{Standard form}: \quad \sf 4x-y=-6[/tex]

Step-by-step explanation:

Given information:

Point-slope form of linear equation:  

[tex]\sf y-y_1=m(x-x_1)[/tex]

(where m is the slope and (x₁, y₁) is a point on the line)

Substitute the given slope and point into the formula:

[tex]\implies \sf y-2=4(x-(-1))[/tex]

[tex]\implies \sf y-2=4(x+1)[/tex]

Slope-intercept form of a linear equation:

[tex]\sf y=mx+b[/tex]

(where m is the slope and b is the y-intercept)

Substitute the given slope and point into the formula and solve for b:

[tex]\implies \sf 2=4(-1)+b[/tex]

[tex]\implies \sf b=6[/tex]

Therefore:

[tex]\sf y=4x+6[/tex]

Standard form of a linear equation:

[tex]\sf Ax+By=C[/tex]

(where A, B and C are constants and A must be positive)

Rearrange the found slope-intercept form of the equation into standard form:

[tex]\implies \sf y=4x+6[/tex]

[tex]\implies \sf 4x-y+6=0[/tex]

[tex]\implies \sf 4x-y=-6[/tex]

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It's common to hear people say that the Renaissance and Reformation saw numerous significant developments.

What impact did the Renaissance have on the Reformation?

Why was the Renaissance important?

The Renaissance period cultivated a new change in art, knowledge, and culture.

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Answer:

Since April 1, 1986 was a Tuesday, March,

1986 had? Mondays.

og

the answer A

Answer:

B, cos(0)=1

Step-by-step explanation:

cos(0)=1

Coordinate A is at point (1,0)

Answer:

Step-by-step explanation:

i think it is c

Answer:

A

Step-by-step explanation:

So let's look at each and determine whether they're true or not.

A. It's vertex is at the origin:

   This is true! The reason for this is because you can express a quadratic in vertex form as such: [tex]f(x)=a(x-h)+k[/tex] where (h, k) is the vertex. If you remember the parent function is simply: [tex]f(x)=x^2[/tex] which we can express as: [tex]f(x) = (x-0)+0[/tex] meaning the vertex is at (0, 0). This also intuitively makes sense, the x^2 will only output positive values for real numbers, so when x<0, f(x) is still going to be positive because it's squaring the negative, so that means when you go from -5 to -4, even though you increased x by 1, the f(x) decreases by 1, since it's squaring the value to get a positive value. The lowest value this can output is 0 because 0^2 = 0. This means you vertex is at the origin.

B. It is not a function:

   So the quadratic parent function only outputs 1 value for each input, although it isn't a one-to-one function, meaning that each output isn't unique, but each input still only outputs 1 output, which makes this a function

C. It is not a parabola:

   parabolas are expressed as quadratic equations, so this is false.

D. It is in quadrants ||| and IV:

   This is not true since the parent function opens upwards and has a vertex at (0, 0) so it will only be in quadrant I and II. I'll provide a diagram showing this.

No, 1775 dollars cannot buy 2 large balloons and 16 small balloons if helium costs $1.60 per cubic foot.

Calculating the Costs of Balloons

It is given that,

Helium costs $1.60 per cubic foot.

Volume of one large balloon = 300 cubic feet

Thus, the price of one large balloon = $1.60 × 300

= $480

Volume of one small balloon = 40 cubic feet

Thus, the price of one small balloon = $1.60 × 40

= $64

How much buying 2 large and 16 small balloons costs?

So, as calculated above,

One large balloon costs $480, and one small balloon costs $64

Therefore, the cost of two large balloons = $480 × 2

= $960

Cost of 16 small balloons = $64 × 16

= $1024

Total cost of balloons = $960 + $ 1024

= $1984

Since, $1984>$1775,

Therefore, $1775 cannot buy all those balloons if helium costs $1.60 per cubic foot.

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a. The equivalent expression for cos(4x) in terms of just x is 8cos⁴x - 8cos²x + 1

b. The equivalent expression for sin(4x) in terms of just x is 4sinxcos³x - 2sinxcosx

To find the trigonometric expression for cos(4x)

cos(4x) = cos[2(2x)]

Now from trigonometric identities

cos2Ф = 2cos²Ф - 1

Let 2x = Ф.

So, cos(4x) = 2cos²2x - 1

Also, cos2x = 2cos²x - 1

So, substituting cos2x into cos4x, we have

cos(4x) = 2cos²2x - 1

cos(4x) = 2[2cos²x - 1]² - 1

cos(4x) = 2[4cos⁴x - 4cos²x + 1] - 1

cos(4x) = 8cos⁴x - 8cos²x + 2 - 1

cos(4x) = 8cos⁴x - 8cos²x + 1

So, the equivalent expression for cos(4x) in terms of just x is 8cos⁴x - 8cos²x + 1

To find the expression for sin(4x)

sin(4x) = sin2(2x)

From trigonometric identities sin2Ф = 2sinФcosФ

Let 2x = Ф

So, sin(4x) = sin2(2x)

= 2sin2xcos2x

Since

Substituting these into the equation, we have

sin(4x) =  2sin2xcos2x

sin(4x) = (2sinxcosx)(2cos²x - 1)

sin(4x) = 4sinxcos³x - 2sinxcosx

So, the equivalent expression for sin(4x) in terms of just x is 4sinxcos³x - 2sinxcosx

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The number of tickets sold on Friday were 66 children tickets and 97 adult tickets.

An equation is an expression that shows the relationship between two or more numbers and variables.

The equation of the parabola passing through (0, 0), (-3, 9) and (3, 9) is y = x²

The equation of the parabola with vertex at (4, 22) passing through (0, 6), (8, 6) and (3, 9) is y < -x² + 8x + 6

The solution to the system of equation is the darker region and it contains the point (-1, 6)

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Answer:

(2,9)

Step-by-step explanation:

Got it on FlVS

The expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

A line is the shortest distance between two points and the point where two lines meet is known as an angle

From the figure shown

m<2 = 58 degrees (alternate exterior angle)

Since the sum of angles on a straight line is 180 degrees, hence;

3x-1+<2 + 4x - 10= 180

3x-1 + 58 + 4x - 10 =180

7x + 47 = 180

7x = 180- 47

7x = 133

x = 19

Hence the expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

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Answer:

angle 2=58 degrees

angle 1=4x-10 degrees

Step-by-step explanation:

Rule 1:

1, 3, 7, 15, 31, 63, 127, 255, 511, 1023 etc

Rule 2:

40, 24, 16, 12, 10, 9, 8.5, 8.25, 8.125, 8.0625 etc

Hope this helps!

The first sequence is 1,3,7,15,31 and second sequence is 40,24,16,12,10.

Given that first rule is that multiply by 2 then add 1 starting from 1 and second rule is that divide by 2 then add 4 starting from 40.

We have to find the sequences for each rule.

We have to start from starting point and then apply the rule to find the sequence.

Sequence of rule 1:

Second term of sequence=2*1+1=3

Third term of the sequence=2*3+1=7

Fourth term of the sequence=2*7+1=15

Fifth term of the sequence=2*15+1=31.

Sequence will be 1,3,7,15,31.

Sequence of rule 2:

Second term of sequence=40/2+4=24

Third term of sequence=24/2+4=16

Fourth term of sequence=16/2+4=12

Fifth term of sequence=12/2+4=10

Sequence will be : 40,24,16,12,10.

Hence the first sequence is 1,3,7,15,31 and second sequence is 40,24,16,12,10.

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Using the binomial distribution, it is found that there is a 0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

With 5 shoots, the probability of making at least one is [tex]\frac{211}{243}[/tex], hence the probability of making none, P(X = 0), is [tex]\frac{232}{243}[/tex], hence:

[tex](1 - p)^5 = \frac{232}{243}[/tex]

[tex]\sqrt[5]{(1 - p)^5} = \sqrt[5]{\frac{232}{243}}[/tex]

1 - p = 0.9908

p = 0.0092

Then, with 6 shoots, the parameters are:

n = 6, p = 0.0092.

The probability that at least two of them make it inside the recycling bin is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which:

[P(X < 2) = P(X = 0) + P(X = 1)

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.0092)^{0}.(0.9908)^{6} = 0.9461[/tex]

[tex]P(X = 1) = C_{6,1}.(0.0092)^{1}.(0.9908)^{5} = 0.0527[/tex]

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.9461 + 0.0527 = 0.9988

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9988 = 0.0012[/tex]

0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

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Answer:

33.3

Step-by-step explanation:

The scale factor is 2/3.

The ratio of the surface areas is the square of the scale factor, so the surface area of the smaller cone is 75(2/3)², which is about 33.3 cm²

Answer:

first option

Step-by-step explanation:

two solutions means 2 cutting points

The slope of the line through the points is -3

The points are given as:

(-1, 8) and (3, -4)

The slope is calculated as:

m = (y2 - y1)/(x2 - x1)

So, we have:

m = (-4 - 8)/(3 + 1)

Evaluate the expression

m = -3

Hence, the slope of the line through the points is -3

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The slope of the line that passes through (-1, 8) and (3, -4) is: -3.

Slope of a line (m) = change in y / change in x.

Given the points, (-1, 8) and (3, -4):

Slope (m) = (-4 - 8)/(3 - (-1))

Slope (m) = (-12)/(4)

Slope (m) = -3

Thus, the slope that passes through the given points is: -3.

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