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Basile [38]
3 years ago
9

C is the midpoint of line segment AB Find the value of x, if line segment AB = 5x and line segment AC = 20.

Mathematics
1 answer:
Softa [21]3 years ago
5 0
D.8

A and C are out because they would get you ether lower or equal to 20, which is equal to AC and your trying to find X in AB which is so posed to be the what equals AC+BC.
AB=AC+BC
5x=20+BC
5(8)=20+BC
40=20+BC
40=20+20

AC equals 20. AB will have to equal AC+ CB. If you put in A.4 in AB 5X you will get twenty and that not what you want. You want to get a number that will get you the an answer greater to AC. So 5(8)=40. 
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Given the function h(x)=-x^2-5x+14 determine the average rate of change of the function over the interval -9
Juli2301 [7.4K]

Answer:

The average rate of change of the function over the interval is of 6.

Step-by-step explanation:

Average rate of change:

The average rate of change of a function h(x) over an interval [a,b] is given by:

A = \frac{h(b)-h(a)}{b-a}

In this question:

Over the interval [-9,-2], so a = -9, b = -2, b - a = -2 -(-9) = 7

The function is:

h(x) = -x^2 - 5x + 14

h(-9) = -9^2 -5(-9) + 14 = -81 + 45 + 14 = -22

h(-2) = -2^2 -5(-2) + 14 = -4 + 10 + 14 = 20

Then

A = \frac{h(-2)-h(-9)}{7} = \frac{20-(-22)}{7} = \frac{42}{7} = 6

The average rate of change of the function over the interval is of 6.

8 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
2 years ago
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