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gtnhenbr [62]
2 years ago
10

F(x)=(x+1)(x-5) find all the critical points ​

Mathematics
1 answer:
adoni [48]2 years ago
4 0

I don't know what you mean by "critical points". However, I'm going to list all the important points in this equation-

x intercepts: (-1 ,0) and (5,0)

vertex: (2,-9)

y intercept: (0,-5)

Hope this helps!

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Sara goes on a slingshot ride in an amusement park. she is strapped into a spherical ball that has a radius of centimeters. what
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I hope this helps you




volume of sphere =4/3.pi.r^3
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3 years ago
Michael and Imani go out to eat for lunch. If their food and beverages cost $25.30 and there is an 8% meals tax, how much is the
blondinia [14]

Answer:

The bill total is $27.32

Step-by-step explanation:

how to figure out tax amount: 25.30 x .08 = $2.024

add tax to cost of meal : 25.30 + 2.024 = $27.32

5 0
3 years ago
The formula to find the spring constant, , for a spring with potential energy, , that has been stretched by length, , is shown b
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the potential energy of the spring

5 0
3 years ago
Read 2 more answers
10
Novay_Z [31]

Answer:

The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the <u>line y = x</u> and a translation <u>10 units right and 4 units up</u>, equivalent to T₍₁₀, ₄₎

Step-by-step explanation:

For a reflection across the line y = -x, we have, (x, y) → (y, x)

Therefore, the point of the preimage A(-6, 2) before the reflection, becomes the point A''(2, -6) after the reflection across the line y = -x

The translation from the point A''(2, -6) to the point A'(12, -2) is T(10, 4)

Given that rotation and translation transformations are rigid transformations, the transformations that maps point A to A' will also map points B and C to points B' and C'

Therefore, a sequence of transformation maps ΔABC to ΔA'B'C'. The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the line y = x and a translation 10 units right and 4 units up, which is T₍₁₀, ₄₎

5 0
3 years ago
What is the value of b ?
HACTEHA [7]

Two polynomials are equal if the coefficients are equal.

x^3+(3+b)x^2+(3b-2)x-6\\\downarrow\qquad\ \ \downarrow\qquad\qquad\ \ \ \ \downarrow\\x^3\ +\ \ 6x^2\ \ \ \ +\ \ \ \ \ 7x\ \ -\ 6\\\\x^3\to1=1\\x^2\to3+b=6\\x\to3b-2=7\\constans\to-6=-6\\\\3+b=6\ \wedge\ 3b-2=7\\\\3+b=6\ \ \ |-3\\b=3\\\\3b-2=\ \ \ \ |+2\\3b=9\ \ \ \ |:3\\b=3\\\\Answer:\ \boxed{b=3}\to\boxed{B.}

6 0
3 years ago
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