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3 years ago

What point is the vertex for the graph of y = |x| + 2?

1 answer:

7 0

The vertex of $y=|x|$ is $(0,0)$ . The vertex of $y=|x|+2$ is the vertex of $y=|x|$ moved 2 units up. So its $(0,0+2)=\boxed{(0,2)}$

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The following table shows the length and width of two rectangles:

faltersainse [42]

The perimeter of these can be found by adding length and width and then multiplying by two.

Rectangle A: (with all of the variables already doubled)
2x + 16 + 2x - 2
2x + 2x + 16 - 2
4x + 14
So rectangle A's perimeter is 4x + 14.

Rectangle B: (Still with all the variables already doubled)
8x + 10 + 6x - 4
8x + 6x + 10 - 4
14x + 6
So rectangle B's perimeter is 14x + 6

And now to subtract the two.
(7x + 3) - (4x + 14)
14x + 6 - 4x - 14
14x - 4x + 6 - 14
10x - 8

So it would be C.

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4 years ago

Choose the system of inequalities that best matches the graph below.

mezya [45]

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3 years ago

Read 2 more answers

Please help!<br>What is the longest side of triangle TUV?

xxMikexx [17]

I say UV is the biggest thanks to my vision lol; and cuz I used a ruler.

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3 years ago

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DaniilM [7]

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3 years ago

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

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3 years ago