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yulyashka [42]
3 years ago
9

Need this ASAP please

Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0
The second answer since f goes inside g which gives u 5x^2
You might be interested in
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
3 years ago
Three boxes weighing 128 pounds each and one box weighing 254 pounds were loaded onto the back of an empty truck. A crate of app
Firlakuza [10]

Answer:

1352 pounds

Step-by-step explanation:

128+128=256

256+254=510

510+128=638

2000-638=1352

6 0
3 years ago
A tree 12m high is broken by the wind in such a way that its top touches the ground and makes an angle 60° with the ground. at w
Katena32 [7]

The height of broken part of tree from ground is 5.569m.

Justification:

Let BD is a tree of height 12 m.

<u>Suppose it got bent at a point C and let the part CD take the position CA, meeting the ground at A</u>.

i.e., CD = AC = h m

<u>Broken part makes 60° angle from ground</u>

So, ∠BAC = 60°

<u>Now, height of remaining part of tree</u> = (12 – h)m.

In right angled ∆ABC,

sin 60° = BC/AC

⇒ √3/2 = (12 - h)/h

⇒ √3h = 2(12 – h)

⇒ √3h = 24 – 2h

⇒ √3h + 2h = 24

⇒ h(√3 + 2) = 24

⇒ h(1.732 + 2) = 24

⇒ h(3.732) = 24

⇒ h = 24/3.732 = 6.4308 m

<u>Hence, height of broken tree from ground</u>

⇒ BC = 12 – h

⇒ 12 – 6.4308 = 5.569m

<u>Hence, tree is broken 5.569 m from ground</u>.

<u>Note</u>: See attached picture.

4 0
3 years ago
Can someone please help me with this plz only on 13
BaLLatris [955]

Answer:

2/9

Step-by-step explanation:

To determine the fraction of students that walk to school, take the fraction of students that do not ride the bus and multiply by the fraction that walk

6/15* 5/9

Simplify the fractions

6/15 Divide the top and bottom by 3

2/5

2/5 * 5/9 = 2/9

4 0
3 years ago
Aiden owns a food truck that sells tacos and burritos. He sells each taco for $4.75 and each burrito for $7. Yesterday Aiden mad
nevsk [136]

Number of tacos sold is 65 and number of burritos sold is 40

<h3><u>Solution:</u></h3>

Given that Aiden sells each taco for $4.75 and each burrito for $7

Let the number of tacos sold be "t" and number of burritos sold be "b"

Given that Aiden sold 25 more tacos than burritos

t = b + 25 ---- eqn 1

Also given that yesterday Aiden made a total of $588.75 in revenue

number of tacos sold x cost of each tacos + number of burritos sold x cost of each burritos = 588.75

t \times 4.75 + b \times 7 = 588.75

4.75t + 7b = 588.75  ----- eqn 2

Substitute eqn 1 in eqn 2

4.75(b + 25) + 7b = 588.75

4.75b + 118.75 + 7b = 588.75

11.75b = 470

b = 40

Substitute b = 40 in eqn 1

t = 40 + 25

t = 65

Thus the number of tacos sold is 65 and number of burritos sold is 40

5 0
3 years ago
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