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creativ13 [48]
3 years ago
14

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

Mathematics
1 answer:
Mama L [17]3 years ago
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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What algebraic expression is a polynomial with a degree of 4
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]

That is correct. On APEX its C. g(x)= 1/3 |x|

3 0
3 years ago
Jacob ate 21 chips ahoy cookies in 1 hour and Bryant ate 25 chips ahoy cookies in the same time, at this rate, how mant whole co
marissa [1.9K]
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55 cookies in 80 mins
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4 0
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]
For this case we have the following equation:
 h (t) = - 12t2 + 36t
 When the object hits the ground we have:
 - 12t2 + 36t = 0
 We look for the roots of the polynomial:
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 Therefore, the time it takes the object to hit the ground is:
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 Answer:
 
the time when the object hits the ground is:
 
t = 3 s
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 <em>R</em> is the region (in polar coordinates)

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