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dimulka [17.4K]
4 years ago
10

Help please..I really need it​

Mathematics
1 answer:
jarptica [38.1K]4 years ago
4 0

Answer:

256

Step-by-step explanation:

V=1/3*π*4^2*7,46=124.993

32000/124.99=~256

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SOMEONE PLEASE HELP ME WILL GIVE BRAINLY
Vitek1552 [10]

Answer:

Let me get my homie on here to help you.

Step-by-step explanation:

3 0
3 years ago
A train travels at a speed of 90 miles per hour. How far will it have traveled<br> in 3 hour
DiKsa [7]
270 miles be 90+90+90 equals 270 and thats the answer
5 0
3 years ago
Read 2 more answers
Find the indicated values, where g(t) = t^2 - t and f(x) = 1 + x g(f(3)-2f(1))
VashaNatasha [74]

Answer:

B. 0

Step-by-step explanation:

Given

g(t) = t² - t

f(x) = 1 + x

Required

Find g(f(3) - 2f(1))

First, we'll solve for f(3)

Given that f(x) = 1 + x

f(3) = 1 + 3

f(3) = 4

Then, we'll solve for 2f(1)

2f(1) = 2 * f(1)

2f(1) = 2 * (1 + 1)

2f(1) = 2 * 2

2f(1) = 4

Substitute the values of f(3) and 2f(1) in g(f(3) - 2f(1))

g(f(3) - 2f(1)) = g(4 - 4)

g(f(3) - 2f(1)) = g(0)

Now, we'll solve for g(0)

Given that g(t) = t² - t

g(0) = 0² - 0

g(0) = 0 - 0

g(0) = 0

Hence, g(f(3) - 2f(1)) = 0

From the list of given options, the correct answer is B. 0

5 0
3 years ago
03.03 mc) find the ordered pairs for the x- and y-intercepts of the equation 5x − 6y = 30 and select the appropriate option belo
Archy [21]
5x - 6y = 30

to find x intercept, sub in 0 for y
5x - 6(0) = 30
5x = 30
x = 30/5
x = 6......so the x intercept is (6,0)

to find the y intercept, sub in 0 for x
5(0) - 6y = 30
-6y = 30
y = 30/-6
y = -5...so the y intercept is (0,-5)
6 0
3 years ago
Y" - 4y = (x2 - 3) sin 2x
Zolol [24]
y''-4y=0

has characteristic equation

r^2-4=0

which has roots at r=\pm2, giving the characteristic solution

y_c=C_1e^{2x}+C_2e^{-2x}

For the nonhomogeneous part of the ODE, let y_p=(a_2x^2+a_1x+a_0)\sin2x+(b_2x^2+b_1x+b_0)\cos2x. Then

{y_p}''=(-4b_2x^2+(8a_2-b_1)x+4a_1-4b_0+2b_2)\cos2x+(-4a_2x^2+(-4a_1-8b_2)x-4a_0+2a_2-4b_1)\sin2x

Substituting into the ODE gives

(-8b_2x^2+(8a_2-b_1)x+4a_1-8b_0+2b_2)\cos2x+(-8a_2x^2+(-8a_1-8b_2)x-8a_0+2a_2-4b_1)\sin2x=(x^2-3)\sin2x

It follows that

\begin{cases}-8b_2=0\\8a_2-8b_1=0\\4a_1-8b_0+2b_2=0\\-8a_2=1\\-8a_1-8b_2=0\\-8a_0+2a_2-4b_1=-3\end{cases}\implies\begin{cases}a_2=-\dfrac18\\\\a_1=0\\\\a_0=\dfrac{13}{32}\\\\b_2=0\\\\b_1=-\dfrac18\\\\b_0=0\end{cases}

which yields the particular solution

y_p=-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x

So the general solution is

y=y_c+y_p
y=C_1e^{2x}+C_2e^{-2x}-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x
4 0
3 years ago
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