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3 years ago

• To round a number, look at the digit to the right of the

Mathematics

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

2. A) 37 B)160 C) 110

The tens digit is 10,20,30,40,50,60,70,80,90. Two away to the left of a decimal point.

3. A)730 B) 675 C)1930

The hundreds digit is three from the left of a decimal point.

4. A) 0.40 B)74.60 C)15.20

The tenth is one away to the right of the decimal.

5. A)1.210 B) 3.430 C)0.350

The hundreth is two away to the right of the decimal point.

Explanation:

If it is above 5 you bring the digit before it up by one, if it is under 5 you bring the digit that is under 5 to zero.

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Find |37|. 0 -37 37 i have 3 minutes to finish HELPP :(

valentina_108 [34]

It’s -37! | this means the opposite of the number ;)

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2 years ago

How many 2 inch segments are there in 12 ft

ollegr [7]

The correct answer for this question is that there are 72 2-inch segment in 12ft. Hopefully that answered your question for you!

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3 years ago

Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$

$V = \int \limits^{\pi/2}_{\pi/4} \ sin \phi \ d \phi \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho ^2 d\rho$

$V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}$

$V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]$

V = 243 $\sqrt{2} \ \pi$

4 0

3 years ago

Which expression does NOT simplify to 1?

koban [17]

fg^3 / g^3f = 1

f^4 / f*f*f*f = f^4 / f^4 = 1

f^2g^3 / f^3g^2 = g/f

f*f/f^2 = f^2 / f^2 = 1

Answer:

f^2g^3 / f^3g^2

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3 years ago

Read 2 more answers

-4(j+k+g) plz help using box method

disa [49]

Answer:

-4j-4k-4g

Step-by-step explanation:

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3 years ago