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4 years ago

Which fraction is equal to 5/6? 8/22, 25/30, 5/15?

1 answer:

3 0

If the fraction cannot be simplified any more, you can multiply both the numerator and denominator by an number.

5/6 (multiply the top and bottom by 2) is the same as 10/12
8/22 (can be simplified by dividing top and bottom by 2) is 4/11
25/30 (simplified by dividing by 5) is 5/6
5/15 (simplified by dividing by 5) is 1/3

Hope this helps!

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8=4w/3, if using this expression what is w equal to?

pogonyaev

8 = 4w/3 Original Problem.
4w/3 = 8 Flip Flopped the problem around.
4w = 24 Multiplied each side by 3.
w = 6 Divided each side by 4.

Answer: w = 6

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4 years ago

Read 2 more answers

Identify the variation as direct, inverse, joint or combined. y = cx²

viva [34]

Answer:

combined variation

Step-by-step explanation:

we have

$y=cx^{2}$

we know that

a) in the interval ----> (0,∞)

If the value of x increase the value of y increase

Is a direct variation

b) in the interval ----> (-∞,0)

If the value of x increase the value of y decrease

Is a inverse variation

therefore

Is a combined variation

4 0

3 years ago

Read 2 more answers

I need help with 1-2 and 4-6!!!

serious [3.7K]

Answer:

1. y = 1/4 and y = 1/4x+2

2. y - 1 = -x (x + 7) and y = -x and x + y = 3

-------

4. y = 3 and x = 6

5. y = 1/2x + 2 and y + 1 = -2x

6. Slope of AB = 2/3

Slope of BC = -3/2

AB is perpendicular to BC because 2/3(-2/3) = -1

ABC is a right triangle because it contains a right angle

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3 years ago

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Zinaida [17]

Answer:

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Step-by-step explanation:

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3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago