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Dahasolnce [82]
3 years ago
9

Identify the y-intercept (initial value) in the function

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer: 13

Step-by-step explanation:

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Triangle STV has vertices S (-3, -2) , T (-4, 3) and V (-2, 3). If (x, y) arrow (x+2, y-3), what are the vertices of its image?
jeka94

Answer:

  see below

Step-by-step explanation:

The mapping tells you how to find the vertices of the image:

  (x, y) ⇒ (x +2, y -3)

  S(-3,-2) ⇒ S'(-3 +2, -2 -3) = S'(-1, -5) . . . . . matches choice D

  T(-4, 3) ⇒ T'(-4 +2, 3 -3) = T'(-2, 0)

  V(-2, 3) ⇒ V'(-2 +2, 3 -3) = V'(0, 0)

8 0
3 years ago
The coordinates of point 7 are given. The midpoint of ST is (5. - 8). Find the coordinates of point S.
Mamont248 [21]

Answer:

S(0, 2)

Step-by-step explanation:

Midpoint Formula: (\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )

Step 1: Plug in known variables

5 = (10 + x)/2

-8 = (18 + y)/2

Step 2: Solve

10 = 10 + x

x = 0

-16 = 18 + y

y = 2

Step 3: Write coordinates

(0, 2)

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Triangle A is the pre-image in a translation, and triangle B is the image, as shown below.
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Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th
STatiana [176]

We can find the acceleration via

{v_f}^2-{v_i}^2=2a\Delta x

We have

\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)

\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}

Then by definition of average acceleration,

a_{\rm ave}=\dfrac{v_f-v_i}t

so that

-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t

\implies\boxed{t=6.33\,\mathrm s}

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2

Then

\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2

\implies\boxed{t=6.33\,\mathrm s}

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