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2 years ago

7

the parent function of the function g(x)=(x-h)^2+k is f(x)=x^2. the vertex of the function g(x) is located at (9,8). what are th

e values of h and k?

1 answer:

Lunna [17]2 years ago

8 0

Answer:

<h2>h = 9, k = 8</h2>

Step-by-step explanation:

$\text{The vertex form of an an equation of a quadratic function:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ g(x)=(x-h)^2+k,\ \text{and the vertex in}\ (9,\ 8).\\\\\text{Therefore}\ h=9\ \text{and}\ k=8.\ \text{substitute:}\\\\g(x)=(x-9)^2+8$

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Answer:

A

Step-by-step explanation:

Solving for 1995, which is 5 years after 1990, so plug in 5 as t.

n= 1.76(5)² - 13.32(5) + 41

n= 44 - 66.6 + 41

n= 18.4

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2 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

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Answer:

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iren2701 [21]

For question 1:
1% = 120 ÷ 100 = 1.2
15% = 1.2 × 15 = 18
So, 15% of 120 is 18

For question 2:
68.4 ÷ 72 = 0.95
0.95 × 100 = 95%
So, the percentage is 95%

For question 3,
1% = 12.5 ÷ 8 = 1.5625
100% = 1.5625 × 100 = 156.25
So, 8% of 12.5 is 156.25.

For question 4,
1% = 48 ÷ 40 = 1.2
100% = 1.2 × 100 = 120
So, 40% of 48 is 120.

For question 5,
1% = 195 ÷ 100 = 1.95
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