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3 years ago

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What are all natural number factors of 130

1 answer:

harina [27]3 years ago

3 0

<span>1, 2, 5, 10, 13, 26, 65, 130 are the natural number factors of 130.

</span>

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A spherical fish bowl is half-filled with water. T

olganol [36]

The volume of a sphere of radius r is given by

$V=\dfrac{4}{3}\pi\cdot r^3$

For your given conditions, the volume of 1/2 a sphere becomes

$V=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{22}{7}\cdot 8^3$

Seems to match your 2nd choice:

... 1 over 24 over 322 over 7(83)

7 0

4 years ago

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Whats the inverse of 17=1+9n

WINSTONCH [101]

Answer:

$\frac{9}{16}$

Step-by-step explanation:

$17 = 1 + 9n$

Subtract 1 from both sides

$17 - 1 = 1 - 1 + 9n$

$16 = 9n$

Divide through by 9

$\frac{16}{9} = \frac{9n}{9}$

$n = \frac{16}{9}$

The inverse of the answer is

$\frac{9}{16}$

7 0

3 years ago

PLEASE HELP WITH THIS

Anuta_ua [19.1K]

Answer:

$528\ cm^2$

Step-by-step explanation:

1. Approach

The surface area is the two-dimensional space around a three-dimensional object. In other words, if one was going to wrap an object with paper, the surface area is the amount of paper one would need. To find the surface area, one could find the area of each face of the prism and then add up all of the values.

2. Finding the surface area,

To find the area of each face of the prism, one must multiply the length of the face by the width of the face. After doing so, add up all of the resulting values.

10 * 20 = 200

8 * 20 = 160

6 * 20 = 120

$\frac{1}{2}$ * 6 * 8 = 24

$\frac{1}{2}$ * 6 * 8 = 24

Add up all of the values,

200 + 160 + 120 + 24 + 24

= 528

4 0

3 years ago

Someone Help me please !

EastWind [94]

Answer:

$\sqrt{9} \times \sqrt{16}$

Step-by-step explanation:

$\sqrt{9} \times 16 = \sqrt{9} \times \sqrt{16} = 3 \times 4 = 12$

Hope this helps ;) ❤❤❤

6 0

3 years ago

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Solve using identities

jeka57 [31]

Answer:

Solution given

Cos $\displaystyle \theta_{1}=\frac{13}{15}$

consider Pythagorean theorem

$\bold{Sin²\theta+Cos²\theta=1}$

Subtracting $Cos²\theta$ both side

$\displaystyle Sin²\theta=1-Cos²\theta$

doing square root on both side we get

$Sin\theta=\sqrt{1-Cos²\theta}$

Similarly

$Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}$

Substituting value of $Cos\theta_{1}$

we get

$Sin\theta_{1}=\sqrt{1-(\frac{-13}{15})²}$

Solving numerical

$Sin\theta_{1}=\sqrt{1-(\frac{169}{225})}$

$Sin\theta_{1}=\sqrt{\frac{56}{225}}$

$Sin\theta_{1}=\frac{\sqrt{56}}{\sqrt{225}}$

$Sin\theta_{1}=\frac{\sqrt{2*2*14}}{\sqrt{15*15}}$

$Sin\theta_{1}=\frac{2\sqrt{14}}{15}$

Since

In III quadrant sin angle is negative

<h3> $\bold{Sin\theta_{1}=-\frac{2\sqrt{14}}{15}}$ </h3>

3 0

3 years ago

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