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3 years ago

Khan academy: Point C is the image of C (-4,-2) under a reflection across the y-axis.

Mathematics

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

The coordinates of point C' are (4, -2)

here's why:

When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is taken to be the additive inverse. The reflection of point (x, y) across the y-axis is (-x, y). So if we use this information with the original coordinates, (-4, -2), the new coordinates will be (4, -2).

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Use inverse operations to complete the second equation each time

ratelena [41]

Answer:

a) 55-42=13

b) 29-10=19

c) 65÷5=13

d) 72÷8=9

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3 years ago

Subtract 5x^2 – 2x – 5 from – 2x^2 - 5x - 7

user100 [1]

Answer:

$-7x^{2} -3x-2$

Step-by-step explanation:

$(-2x^2-5x-7) - (5x^2-2x-5)\\\\-2x^2 - 5x^2 = -7 x^2\\-5x - -2x = -3x\\-7 - -5 = -2 \\Result: -7x^2-3x-2$

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3 years ago

Read 2 more answers

H(-4), h(0), h(4), h(8)

DIA [1.3K]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The first two x-values, -4 and 0, fall in the first part of the function definition.

There is no function definition for x = 4, as the second part only applies for x > 4.

The last x-value, 8, falls in the second part of the function definition.

The function is evaluated by putting the x-value where x is in the expression and doing the arithmetic. For example, ...

h(-4) = -(-4+1)^2 +7 = -(-3)^2 +7 = -9 +7 = -2

h(0) = 6

h(4) = undefined

h(8) = -8/3

6 0

3 years ago

Each day of overtime you work, you earn an additional $50 more than your regular pay of $100 a day.If you worked overtime 3 out

shusha [124]

The answer is 650 because 100 * 5 = 500 and 50 * 3 = 150 and 500 + 150 = 650

3 0

4 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

3 years ago