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malfutka [58]
3 years ago
7

Increase #500 in the ratio 16:10

Mathematics
1 answer:
Oxana [17]3 years ago
3 0

Answer:

Answer is <em>900</em>.

Step-by-step explanation:

To find:

Increase #500 in the ratio 16:10

Solution:

<em>New Number: Old Number = 16:10</em>

We are given the old number as 500.

Let the new number after increase = x

Now, using the above ratio:

<em />x<em>: </em>500<em> = </em>16:10

\dfrac{x}{500} = \dfrac{16}{10}\\\Rightarrow x = 500 \times \dfrac{16}{10}\\\Rightarrow x = 50 \times 16\\\Rightarrow x = 900

Therefore, the increased value of 500 in the ration 16:10 is <em>900</em>.

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3

Step-by-step explanation:

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A young sumo wrestler decided to go on a special diet to gain weight rapidly. He gained weight at a constar
Masteriza [31]

Answer:

  • <u>5.2 kg per month.</u>

<u></u>

Explanation:

The measure of how fast the wrestler gained weight, in kg per month, is the  weight gain rate.

The table is:

Time (months)           Weight (kg)

    0.5                             80.6

    2                                 88.4

    3.5                              96.2

The weight gain rate, WGR, for a period may be calculated as:

  • WGR = [ weight gain ] / [period]

  • WGR = [final weight - initial weight] / [final time - initial time]

  • WGR₁ = [88.4 kg - 80.6 kg] / [2 month - 0.5 month] = 7.8 kg / 1.5 month = 5.2 kg / month

  • WGR₂ = [96.2 kg - 88.4 kg] / [3.5 month - 2 month] = 7.8 kg / 1.5 month = 5.2 kg / month.

As you see, the wrestler gained weight at a constant rate of 5.2 kg per month.

7 0
3 years ago
A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0
3 years ago
Answer Both Questions
labwork [276]

Answer:is the first answer 15.875 and the second answer 17 x 28 ÷5

Step-by-step explanation:

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Point B is the midpoint of Line segment A C .
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Answer:

Angle ABC is bisected by BD

BC =Half AC

2 mangle DBC =mangle ABC

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