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3 years ago

Find the Volume of the square pyramid of the area of the base is 48cm and the height is 8cm.

Mathematics

2 answers:

Over [174]3 years ago

8 0

Answer:

Step-by-step explanation:

too much explaining but trust me

Katen [24]3 years ago

7 0

Answer:128cm^3

Step-by-step explanation:

Base area=48cm^2

Height=8cm

Volume=1/3 x base area x height

Volume=1/3 x 48 x 8

Volume=(1x48x8)/3

Volume=384/3

Volume =128cm^3

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Need help with this math problem

AysviL [449]

Answer:

CAB = 54

ABC = 21

ACB = 105

DCB = 75

Step-by-step explanation:

18x - 15 = (13x - 11) + (4x + 1)

18x - 15 = 17x - 10

x = 5

CAB = 13x - 11 = 13(5) - 11 = 65 - 11 = 54

ABC = 4x + 1 = 4(5) + 1 = 20 + 1 = 21

ACB = 180 - (CAB + ABC) = 180 - (54 + 21) = 180 - 75 = 105

DCB = 180 - ACB = 180 - 105 = 75

3 0

4 years ago

What is the kinetic energy of a 50-kg child running to catch the school bus at 2 m/s

aev [14]

Answer:

100 J

Step-by-step explanation:

The formula for kinetic energy is as follows:

$\boxed{KE= \frac{1}{2} mv²}$

where

KE= kinetic energy (J)

m= mass (kg)

v= velocity (m/s)

Given: m= 50 kg, v= 2 m/s

Substituting the given information into the formula:

= ½(50)(2²)

= ½(50)(4)

= 100 J

7 0

3 years ago

3. Which number has the greatest absolute value? A)|-3| B)|-8| C)|2| D)|5|

Bond [772]

Answer:

Step-by-step explanation:

|-8|=8, and all the others are less. hope that helped

7 0

3 years ago

Please help!! I don’t know what to do

ira [324]

Answer:

x = 14.4

Step-by-step explanation:

Similar means that the figures are proportional to each other. Because of this, we can form a problem. $\frac{9}{5}$ (the short side lengths) = $\frac{x}{8}$ (the long side lengths). Now we can solve this by cross-multiplying. If we multiply 9 · 8 we get 72, and 5 · x is 5x. 72 = 5x. Now divide both sides by 5. 72 ÷ 5 = 14.4. Therefore, x should be equal to 14.4. Does this make sense?

6 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago