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yuradex [85]
3 years ago
13

What is the perimeter of a square whose diagonal is 10 centimeters? Round to the nearest tenth, if necessary.

Mathematics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

D

Step-by-step explanation:

I hope this correct and have a great day

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56 1/4% in simplest form. don't understand how to do it
Illusion [34]
I'd turn the fraction into a decimal. 56.25% but the answer in simplest form is <span>-5/7(2/5). i hope this helped you. :)</span>


5 0
3 years ago
Read 2 more answers
Solve x^2-6x+25 = 0 by completing the square.
Oksana_A [137]

First, subtract both sides by 25:

x^2 - 6x = -25

Because we're completing the square, we subtract both sides by 9:

x^2 - 6x - 9 = -34

(x - 3) = 5.831 (about)

x = 8.831

3 0
3 years ago
Joe is saving money for his
tatyana61 [14]

Answer:

He will need to save $295

Step-by-step explanation:

he needs 450 and he has 155, so all you would do is subtract 155 from 450 and you get the rest he needs which is 295

6 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
Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? W
Anna71 [15]

Answer: The resultant would be the sum and the difference between the vectors.

Step by step explanation: 1. The possible resultant is between the sum of the 2 vectors and the difference between the two vectors.

2. The greatest magnitude is when the vectors lie in the same direction and the sum would be the scalar sum of the two vectors. The angle between the two would be zero degree.

6 0
3 years ago
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