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SOVA2 [1]
2 years ago
15

Mrs Atkins is going to choose two students from her class to take part in a competition.

Mathematics
1 answer:
Schach [20]2 years ago
6 0

Given:

Total number of girls in her class = 16

Total number of boys in her class = 14

To find:

The number of different ways of choosing one girl and one boy.

Solution:

We have,

Total number of girls = 16

Total number of boys = 14

So,

Total number of ways to select one girl from 16 girls = 16

Total number of ways to select one boy from 14 boys = 14

Now, number of different ways of choosing one girl and one boy is

16\times 14=224

Therefore, the required number of different ways is 224.

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Triss [41]
The answer is 41 inches. Since <span>△HKS∼△DLY, so the measurements of the angles are just the same. So if the sides are   15 x 20 x 32. Then to get the sides of the other triangle is simply adding what is the measurement of the sides from the first triangle. so it is 24 x 29 x 41. So the answer of DY is equal to 41 inches.</span>
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2 years ago
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When finding the volume of a soup can, Tommy needs to calculate the area of the base. Describe what type of base Tommy’s soup ca
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volume=[base area]*height
volume=πr^2h
where;
base area=πr^2
height=h

3 0
2 years ago
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The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
1 year ago
Torrey starts a new job with an annual salary of $60,000. For each year she continues to work for the same company, she will rec
GarryVolchara [31]
The third choice will work
6 0
1 year ago
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Describe in words the region of double-struck R3 represented by the inequality. 0 ≤ z ≤ 7 The inequality 0 ≤ z ≤ 7 represents al
qwelly [4]

Answer:

All points between z = 0 and z = 7 along the z-azis in R3 x-y-z plane

Step-by-step explanation:

The inequality 1 \le z  \le7 represents all sets of points on the z-axis in the R3 plane that lies bewteen z = 0 and z= 7.

It is represents a line segment joining two point z= 0 and z=7 along z- axis in the R3 plane, while x=0 and y=0 on the x-y plane.

4 0
2 years ago
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