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3 years ago

What is the image of (0,7) after a reflection over the x-axis

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

7 0

Answer:

(0, - 7 )

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y ) → (x, - y ), thus

(0, 7 ) → (0, - 7 )

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Find (f•f) (0) F(x)=3x-2

NNADVOKAT [17]

Answer: (f·f)(0) = 4

(fof)(0) = -8

Step-by-step explanation:

f(x) = 3x - 2

(f·f)(x) = (3x - 2)(3x - 2)

= 9x² - 12x + 4

(f·f)(0) = 9(0)² - 12(0) + 4

= 4

(fof) = 3(3x - 2) - 2

= 9x - 8

(fof)(0) = 9(0) - 8

= -8

I wasn't sure if you wanted multiplication or composition so I solved both

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3 years ago

Larry,Moe,and tome at a pizza together. If larry ate 3/8 of the pizza, and tom ate 1/6 of it, how much did moe eat?

mariarad [96]

3/8 + 1/6 is 13/24. 24/24 -13/24 is 11/24.Moe ate 11/24.

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3 years ago

Help on #26 please. State the domain and range

ladessa [460]

Wouldnt the domain be x with the equal cross thingy 3

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3 years ago

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$

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3 years ago

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azamat

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