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BartSMP [9]
4 years ago
5

5/8 divided by 1 1/3

Mathematics
2 answers:
ElenaW [278]4 years ago
8 0

(5/8) / (1 1/3) = 0.46875 in decimal form

And 15/32 in fraction form

lesya692 [45]4 years ago
5 0

Answer: 15/32

Step-by-step explanation:

To solve, 5/8 ÷ 1 1/3

First convert the improper fraction to proper fraction, 1 1/3= 4/3

Then rewrite the question as:

5/8 ÷ 4/3

Change the sign to multiplication and find the reciprocal of the second fraction:

5/8 × 3/4

Multiply the numerators: 5 x 3 = 15

Multiply the denominators: 8 x 4 = 32

= 15/32

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Step-by-step explanation:

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3 years ago
10) Megan's started a service club at her high school. After all of the members signed up
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3 years ago
The equation of y= kx and how k=y/x
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That’s the same equation but different format.
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4 years ago
Read 2 more answers
Can someone tell me the main idea for both of the story's yeh-shen and cinderella
Tpy6a [65]

Answer:

yeh-shen:

The theme in the story is that good things happen to good people and bad things befall on evil people.

cinderella:

Cinderella teaches the morals of kindness towards all, forgiving others for doing wrong, and never letting bad things ruin your heart. The themes of the story are good versus evil and luck changing your life.

Step-by-step explanation:

5 0
2 years ago
The length of a rectangle is (2x –1) units, and the width is (x − 2) units. The area of the rectangle is 20 square units. What i
DedPeter [7]

Length of the given rectangle is 8 units.

Option - D

<u>Solution:</u>

Given that  

Length of a rectangle = (2x-1)\text { units}

Width of a rectangle = (x-2)\text{ units}

Area of rectangle = 20 square units  

Need to calculate length of rectangle.

\text { Formula of area of rectangle }=\text { Length of a rectangle } \times \text { Width of a rectangle }

Substituting given value of Area of rectangle and expressions for length and width of rectangle in formula, we get

20=(x-2)(2 x-1)

On solving above equation for x we get

\begin{array}{l}{20=x(2 x-1)-2(2 x-1)} \\\\ {\Rightarrow 20=2 x^{2}-x-4 x+2} \\\\ {\Rightarrow 2 x^{2}-5 x+2-20=0} \\\\ {\Rightarrow 2 x^{2}-5 x-18=0}\end{array}

On splitting the middle term in such a way that product of split terms comes as 2x^2\times -18 = -36x^2 and summation comes as -5x, we get

\begin{array}{l}{\Rightarrow 2 x^{2}+4 x-9 x-18=0} \\\\ {=>2 x(x+2)-9(x+2)=0} \\\\ {\Rightarrow(2 x-9)(x+2)=0} \\\\ {\Rightarrow(2 x-9)=0 \text { or }(x+2)=0} \\\\ {\Rightarrow x=\frac{9}{2} \text { or } x=-2}\end{array}

\begin{array}{l}{\text { When } x=\frac{9}{2}, \text { length }=(2 x-1) \text { units }=\left(2 \times \frac{9}{2}-1\right)=8 \text { units }} \\\\ {\text { When } x=-2, \text { length }=(2 x-1) \text { units }=(2 \times(-2)-1)=-5 \text { units }}\end{array}

As length cannot be negative hence we can ignore negative value.

So length is 8 units.

3 0
3 years ago
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