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3 years ago

How do you write 39/50 as a percentage?

2 answers:

4 0

We can convert the fraction to a denominator of 100, since percents are parts of 100.

39 * 2 = 78
50 * 2 = 100

So we have 78/100 or 78%

dolphi86 [110]3 years ago

3 0

You write it like this: 78%
HOPE I HELPED :D

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Multiply. Write your answer as a fraction in simplest form.<br><br> −1(4/5)=

Cerrena [4.2K]

Answer: -4/5

Step-by-step explanation:

Any positive number/fraction multiplied by -1 will become negative

6 0

2 years ago

1,2,3, or 4 for answer

slava [35]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

H (t) = 2t - 2 g(t) = 4t + 5 Find (h(g(t))

ASHA 777 [7]

Answer:

$h(g(t))=8t+8$

Step-by-step explanation:

We are given the two functions:

$h(t)=2t-2\text{ and } g(t)=4t+5$

And we want to find:

$h(g(t))$

So, by substitution, we acquire:

$h(g(t))=2(4t+5)-2$

Distribute:

$h(g(t))=8t+10-2$

Simplify:

$h(g(t))=8t+8$

5 0

2 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

What is the square root of 124.9924

Thepotemich [5.8K]

The answer is:
11.18
Hope This Helps!!
:)

4 0

3 years ago