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3 years ago

Use compatible numbers to estimate the quotient 358/3

1 answer:

6 0

Compatible numbers are known as numbers that are the closest value to the actual numbers- this makes mental arithmetic to be performed better. The quotient of 358/3 is equal to 119.333 therefore the compatible number of the quotient of 358/3 is 119.

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X-1/2x-1/4x=75 what the answer of this equation?

vekshin1

Answer:

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5 0

4 years ago

In her last race, bergen swam 1,500 meters in 30 minutes.on average, how many meters did she swim per minutes?

Lelechka [254]

What you will need to do is take 1,500 which is the distance by 30 the speed to get 1500/30 which will get you 50 if you want to check your work divide 1500 by 50 and see if you get 30

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4 0

4 years ago

What is 7.13 x 6 is it 54 because i dont know anything about math

igor_vitrenko [27]

42.96 I think

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8 0

3 years ago

Read 2 more answers

An unknown number w is 30 more than an unknown number p. The number w is also p less than 5

AleksandrR [38]

The problem statement gives rise to two equations.
w = 30 + p . . . . . w is 30 more than p
w = 5 - p . . . . . .. w is p less than 5

Add these two equations to get
2w = 35
w = 35/2 = 17.5
p = 5 - w = -12.5

The unknown number w is 17.5.
The unknown number p is -12.5.

7 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers