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aivan3 [116]
2 years ago
6

What is the volume of the right rectangular prism, in cubic centimeters? A prism has a length of 4 centimeters, width of 9 centi

meters, and a height of 10 centimeters
Mathematics
1 answer:
siniylev [52]2 years ago
8 0

Answer:

360

Step-by-step explanation:

length times width of the base then multiply by the height of the prism.

(4)(9)10

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Find the value of x. Leave answers if necessary in the simplest radical form.
shtirl [24]

Answer:

<h2>x = 10√3</h2>

<u>Step-by-step explanation:</u>

cos 45° = base / hypotenuse

1/√2 = base / 15√2

15√2 = base√2

Base, = 15

sin 60° = perpendicular / hypotenuse

hypotenuse = x

√3/2 = 15 / x

x √3 = 30

x = 30 / √3

x = 10√3

6 0
3 years ago
Read 2 more answers
The region bounded by y=(3x)^(1/2), y=3x-6, y=0
Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is y = (3x)^{\frac{1}{2} }

⇒ y^{2} = 3x ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

y^{2} - y - 6 = 0

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = \int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx

= \sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}

= [\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} }  }{3}] - [13.5 - 18 - 6 + 12]

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0
3 years ago
I need HELP !!!!!!!!!!!!!!!!
Annette [7]

Answer:

I think the solution is (p,q)=(-2,-5)

3 0
3 years ago
The formula for the volume of a cube is V = s3, where s is the length of each side. What is the volume of a cube if each side me
Whitepunk [10]
The answer is 1






HHOPE THIS HELPS!!!!!!!
6 0
3 years ago
Look at the two equations:
Dvinal [7]

Answer:

- 3x + 6 = 21

- 3x = 21 - 6

- 3x = 18

- 3x \div  - 3 = 18 \div  - 3

x =  - 6

- 3x + 6 < 21

- 3x  < 21 + 6

- 3x < 27

- 3x \div  - 3 < 27 \div  - 3

x >  - 9

8 0
3 years ago
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