Common factor of 10 and5

The number of kilograms of water in a human body varies directly as the mass of the body. An 87-kg person contains 58 kg of wat

katen-ka-za [31]

Answer:

50 kg water.

Step-by-step explanation:

We have been given that the number of kilograms of water in a human body varies directly as the mass of the body.

We know that two directly proportional quantities are in form $y=kx$ , where y varies directly with x and k is constant of variation.

We are told that an 87-kg person contains 58 kg of water. We can represent this information in an equation as:

$58=k\cdot 87$

Let us find the constant of variation as:

$\frac{58}{87}=\frac{k\cdot 87}{87}$

$\frac{29*2}{29*3}=k$

$\frac{2}{3}=k$

The equation $y=\frac{2}{3}x$ represents the relation between water (y) in a human body with respect to mass of the body (x).

To find the amount of water in a 75-kg person, we will substitute $x=75$ in our given equation and solve for y.

$y=\frac{2}{3}(75)$

$y=2(25)$

$y=50$

Therefore, there are 50 kg of water in a 75-kg person.

3 0

3 years ago

A study was being conducted about how much college students spend on the purchase of their textbooks.

Mandarinka [93]

Answer:

n=4

Step-by-step explanation:

because 4x10=40

5 0

3 years ago

The projected T. rex is about

dolphi86 [110]

Answer: $scale\ factor=230$

Step-by-step explanation:

You need to analize all the information given in the exercise.

You know that the actual height of the T-rex on the projector lens is the following:

$actual\ height=2\ cm$

The height of the projected T-rex is:

$projected\ height=4.6\ m$

Make the conversion from meters to centimeters:

$projected\ heigth=(4.6\ m)(\frac{100\ cm}{1\ m})\\\\projected\ heigth= 460\ cm$

Therefore, the scale factor will be:

$scale\ factor=\frac{projected\ heigth}{actual\ heigth}$

Finally, you must substitute values into the equation and then you must evaluate in order to find the scale factor of the projection. You get that this is:

$scale\ factor=\frac{460\ cm}{2\ cm}\\\\scale\ factor=230$

7 0

3 years ago

The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

4 0

3 years ago

Which statement is true about the difference 2 square root of 7 − square root of 28?

3241004551 [841]

<span> 2 square root of 7 − square root of 28
= 2</span>√7 - √28
= 2√7 - 2√7
= 0

4 0

3 years ago

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