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LUCKY_DIMON [66]
3 years ago
6

Each side of a rectangular prism measures 2 inches, and the height is 10 inches. What is the volume?

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
6 0
V=lwh
evidently, l=w=2

so
h=10
l=2
w=2

v=2*2*10=40

answer is 40 in^3
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dalvyx [7]
0.12 grams of Vitamin C.

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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. what is the pro
AnnyKZ [126]
First two rolls have to be 1-4 that is 2/3 chance twice and the third can be 4or 5
2/3*2/3*1/3 + the chance that the fourth is the 5 or 6. 
2/3*2/3*2/3*1/3

So the solution is : P=2/3*2/3*1/3 + 2/3*2/3*2/3*1/3
7 0
3 years ago
Find the quotient. 12a^3p^4 ÷ -2a^2p
777dan777 [17]

Answer:

-(6ap³)

Step-by-step explanation:

We have to find the quotients of

\frac{12a^{3}p^{4}}{-2a^{2}p}

We take the common 2a² p from the numerator

= \frac{(6ap^{3}(2a^{2}p))}{(-2a^{2}p)}

= -(6ap³)

Therefore, the quotient after division will be -(6ap³)

4 0
3 years ago
Marcos is mixing a solution in science class. He is mixing 3 ounces of food coloring with every 8 ounces of water. If he wants a
jonny [76]

Answer:

With the 24 ounces of water he would have 9 ounces of food coloring.

Step-by-step explanation:

For ever 8 ounces of water there is 3 ounces of food coloring so if there is 24 ounces of water that mean there is 3 times the original starting amount. So 8x3=24 so there is 24 ounces of the water. Since the8 got multiplied by 3 that means the 3 we haven't used yet will get multiplied by 3 as well so we can find the answer. So 3x3=9. That means we would have 9 ounces of food coloring with the 24 ounces of water.

7 0
3 years ago
Integrate e^x(sin(x) cos(x))
Karo-lina-s [1.5K]
I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx

\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}

Looking for I_2:

\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\  I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I

Replacing:

2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}
6 0
3 years ago
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