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2 years ago

14

On a string of lights for every 6 green lights,there are 4 white lights.If their are 30 green lights,how many white lights are t

here?

2 answers:

s344n2d4d5 [400]2 years ago

7 0

You have to do 4 times 6 then add 30 to it i think

kifflom [539]2 years ago

3 0

30÷5=6

4×5=20

for every 30 green lights there are 20 white lights

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GuDViN [60]

Answer:

11:D

12:B

Step-by-step explanation:

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2 years ago

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In the figure below , ABCD is a square . Points are chosen on each pair of adjacent sides of ABCD to form 4 congruent right tria

Lostsunrise [7]

Answer:

5/9 of the area of square ABCD is shaded

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

To find out what fraction of the area of square ABCD is shaded, divide the shaded area by the total area of square ABCD

step 1

Find out the area of square ABCD

The area of a square is

$A=b^{2}$

where

b is the length side of the square

we have

$b=(x+2x)=3x\ units$

so

$A=(3x)^{2}$

$A=9x^2\ units^{2}$

step 2

Find out the area of the 4 congruent right triangles

$A=4[\frac{1}{2}(x)(2x)]=4x^{2}\ units^2$

step 3

Find out the area of the shaded region

The area of the shaded region is equal to the area of square ABCD minus the area of the 4 congruent right triangles

so

$A=9x^2-4x^{2}=5x^{2}\ units^{2}$

step 4

Divide the shaded area by the total area of square ABCD

$\frac{5x^{2}}{9x^{2}} =\frac{5}{9}$

therefore

5/9 of the area of square ABCD is shaded

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2 years ago

Can 3 gallons and 5 tons be expressed as a ratio?

kirill115 [55]

Yes it can, and it can be written as 3:5, 3 to 5, 3/5

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2 years ago

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2 years ago

The areas of two similar triangles are 20m2 and 180m2. The length of one of the sides of the second triangle is 12m. What is the

guajiro [1.7K]

Answer:

The corresponding side of the first triangle is 4 m.

Step-by-step explanation:

Area of the first triangle: A1=20 m^2

Area of the second triangle: A2=180 m^2

Lenght of one of the sides of the second triangle: s2=12 m

Corresponding side of the first triangle: s1=?

A1/A2=(s1/s2)^2

Replacing the known values:

(20 m^2)/(180 m^2)=[s1/(12 m)]^2

Simplifying:

2/18=[s1/(12 m)]^2

Simplifying the fraction on the left side of the equation dividing the numerator and denominator by 2:

(2/2)/(18/2)=[s1/(12 m)]^2

1/9=[s1/(12 m)]^2

Solving for s1: Square root both sides of the equation:

sqrt(1/9)=sqrt{[s1/(12 m)]^2}

sqrt(1)/sqrt(9)=s1/(12 m)

1/3=s1/(12 m)

Multiplying both sides of the equation by 12 m:

(12 m)*(1/3)=(12 m)*s1/(12 m)

Simplifying:

(12 m)/3=s1

4 m=s1

s1=4 m

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