3 years ago

The line with slope negative-4/3 is steeper than the line with slope negative-2

1 answer:

6 0

The bigger absolute value slope will have the steeper line

so u have slopes of -4/3 and -2.....the absolute values of these are 4/3 (or 1 1/3) and 2........2 is larger then 4/3.....so the line with the slope of -2 is steeper then the line with the slope of -4/3......so the statement u wrote is false

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Monica [59]

Answer:

Any expression that has an X with an exponent of 4 and no other higher exponent.

Step-by-step explanation:

The degree of an expression is the highest exponent on an X in the equation.

4 0

3 years ago

If you vertically compress the absolute value parent function, f(x) = |x|, by a factor of 3, what is the equation of the new fun

Anna007 [38]

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3 years ago

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marissa [1.9K]

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3 years ago

The height of a free falling object at time t can be found using the function, h(t) = - 12t2 + 36t. Where h(t) is the height in

never [62]

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Answer:
the time when the object hits the ground is:
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7 0

3 years ago

Read 2 more answers

Tính tích phân sau bằng cách dùng tọa độ cực I=∫∫ <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B

xenn [34]

It sounds like R is the region (in polar coordinates)

R = {(r, θ) : 2 ≤ r ≤ 3 and 0 ≤ θ ≤ π/2}

Then the integral is

$\displaystyle \iint_R\frac{\mathrm dx\,\mathrm dy}{\sqrt{x^2+y^2}} = \int_0^{\pi/2}\int_2^3 \frac{r\,\mathrm dr\,\mathrm d\theta}{\sqrt{r^2}} \\\\ = \int_0^{\pi/2}\int_2^3 \mathrm dr\,\mathrm d\theta \\\\ = \frac\pi2\int_2^3 \mathrm dr \\\\ = \frac\pi2r\bigg|_2^3 = \frac\pi2 (3-2) = \boxed{\frac\pi2}$

5 0

3 years ago