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FrozenT [24]
1 year ago
14

The center of an ice rink is located at (0, 0) on a coordinate system measured in feet. sandi skates along a path that can be mo

deled by the equation y = 0.08x2 - 1.6x 13. david starts at (20, 13) and skates along a path that can be modeled by a quadratic function with a vertex at (10, 5). which system of equations models the paths of the two skaters? y = 0.08x2-1.6x 13 and y = 0.08(x-10)2 5 y = 0.08x2-1.6x 13 and y = 1.6(x-10)2 5 y = 0.08x2-1.6x 13 and y = 0.8(x-10)2 5 y = 0.08x2-1.6x 13 and y = 5(x-10)2 5
Mathematics
1 answer:
WARRIOR [948]1 year ago
5 0

The answers to the question are are the equations

y = 0.08x²-1.6x+13

y = 0.08(x-10)²+5

<h3>How to solve for the system of equations</h3>

David starts at (20, 13) and skates along a path that can be modeled by a quadratic function with a vertex at (10, 5).

Vertex form of quadratic function is

y = a(x-h)^2+k

13 = a(x-10)²+5

let x = 20

a(20-10)²+5

13 = 100a + 5

take like terms

13 - 5 = 100a

8 = 100a

divide through by 100

a = 0.08

hence the equation would be

y = 0.08(x-10)²+5

Read more on equations here:

brainly.com/question/9621134

#SPJ1

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M angle IMJ= help me please thanks
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<h3>How to solve for IMJ?</h3>

To solve for angle IMJ, we make use of the following arc intercept theorem.

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What is 2and1/3 ÷ 1and 1/2 just look at the pic plz help<br><br>A.2 1/6 <br><br>B. 3 1/2
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Find the slope of the line
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Step-by-step explanation:

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6 0
3 years ago
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Please help!!!!!!!!!!!!!!!!!!
elena55 [62]

h(x)=(f\circ g)(x)\\\\h(x)=\sqrt[3]{x+3};\ f(x)=\sqrt[3]{x+2}\\\\h(x)=\sqrt[3]{x+1+2}=\sqrt[3]{(x+1)+2}=\sqrt[3]{g(x)+2}\to \boxed{g(x)=x+1}

-----------------------------------------------------------------------------------------------

f(x)=2x;\ g(x)=2x-1;\ h(x)=\sqrt{x}\\\\(f\circ g\circ h)(9)=?\\\\h(9)=\sqrt9=3\\\\(g\circ h)(9)=g(3)=2(3)-1=6-1=5\\\\(f\circ g\circ h)(9)=f(5)=2(5)=10\\------------------------------\\other\ method:\\\\(f\circ g\circ h)(x)=2(2\sqrt{x}-1)=4\sqrt{x}-2\\\\(f\circ g\circ h)(9)=4\sqrt9-2=4(3)-2=12-2=10

-------------------------------------------------------------------------------------------------

f(x)=4x+1;\ g(x)=x^2-5\\\\(f\circ g)(4)=?\\\\g(4)=4^2-5=16-5=11\\(f\circ g)(4)=f(11)=4(11)+1=44+1=45\\----------------------------------\\other\ method:\\\\(f\circ g)(x)=4(x^2-5)+1=4x^2-20+1=4x^2-19\\\\(f\circ g)(4)=4(4)^2-19=4(16)-19=64-19=45


8 0
3 years ago
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