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Naya [18.7K]
2 years ago
13

Evaluate. Express your answers as whole numbers or fractions.

Mathematics
1 answer:
Yuliya22 [10]2 years ago
8 0

Raise 4 to the power of 4 =256
4 to the -4 = 1/256
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A. (-10,-10)<br> B. (-1 /12, 9 1/2) <br> C. (1/2, 4) <br> D. (1,8)
lara31 [8.8K]

Answer:

I think The answer is ( -10, 10 )

5 0
2 years ago
30% of $40 is blank money
Masja [62]

Answer:

12

Step-by-step explanation:

30% is 0.3,  40 x 0.3 . is 12 :DD

8 0
2 years ago
If Mark gets 150 free miles, how many miles will he be charged for with these two odometer readings: 29,582 and 30,283?
lapo4ka [179]
He will be charged for 551 miles.

30283-29582=701 miles traveled in the rental vehicle.

150 of those miles are free; 701-150 = 551 miles he will be charged for.
8 0
3 years ago
Read 2 more answers
Evaluate using <br> Definite integrals
swat32

Since [0,4]=[0,1]\cup(1,4], we can rewrite the integral as

\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt

Both integrals are quite immediate: you only need to use the power rule

\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}

to get

\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4

Now we only need to evaluate the antiderivatives:

\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15

So, the final answer is 15.

4 0
3 years ago
the initial vertical velocity of a rocket shot straight up is 42 meters per second. How long does it take for the rocket to retu
Snowcat [4.5K]
It takes 4.3 seconds for the rocket to return to earth.

The equation is:
0=-9.8t^2+v_0t+h_0
where -9.8m/sec² is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height.  We will go from the assumption that the rocket is launched from the ground, so h₀=0, and we are told that the initial velocity, v₀, is 42.  This gives us:

0=-9.8t^2+42t

We will use the quadratic formula to solve this.  The quadratic formula is:
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Plugging in our information we have:
t=\frac{-42\pm \sqrt{42^2-4(-9.8)(0)}}{2(-9.8)}&#10;\\&#10;\\=\frac{-42\pm \sqrt{1764-0}}{-19.6}&#10;\\&#10;\\=\frac{-42\pm \sqrt{1764}}{-19.6}&#10;\\&#10;\\=\frac{-42\pm 42}{-19.6}=\frac{-42-42}{-19.6} \text{ or } \frac{-42+42}{-19.6}&#10;\\&#10;\\=\frac{-84}{-19.6}\text{ or }\frac{0}{-19.6}&#10;\\&#10;\\=4.3\text{ or }0

x=0 is when the rocket is launched; x=4.3 is when the rocket lands.
8 0
3 years ago
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