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3 years ago

How do you use the LCM (least common multiple) to write two or more fractions with a common denominator?

1 answer:

7 0

E.g. 1/4 and 1/3
The LCM of the denominators (4 and 3) is 12
So 1/4 can be written as 3/12
And 1/3 can be written as 4/12
You can then add the fractions if you wish to make 7/12

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What is (-2)( 3 4/7) A)-7 1/6 B)-6 6/7 C)-2 3/7 D)-1 2/7

Arte-miy333 [17]

Answer:

Step-by-step explanation:

6 0

2 years ago

the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 58 in, find its length and width

dolphi86 [110]

Answer:

Length = 17 inches

Width = 12 inches

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Step-by-step explanation:

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As it is given that, the length of a rectangle is 5 in longer than its width and the perimeter of the rectangle is 58 in and we are to find the length and width of the rectangle. So,

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Let us assume the width of the rectangle as x inches and therefore, the length will be (x + 5) inches .

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Now, According to the Question :

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${\longrightarrow \qquad { \pmb{\frak {2 ( Length + Breadth )= Perimeter_{(Rectangle)} }}}}$

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${\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}$

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${\longrightarrow \qquad { {\sf{2 ( 2x + 5 )= 58 }}}}$

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${\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}$

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${\longrightarrow \qquad { {\sf{ 4x = 58 - 10}}}}$

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${\longrightarrow \qquad { {\sf{ 4x = 48}}}}$

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${\longrightarrow \qquad { {\sf{ x = \dfrac{48}{4} }}}}$

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${\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\: \: \bigstar$

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Therefore,

The width of the rectangle is 12 inches .

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Now, We are to find the length of the rectangle:

${\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}$

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${\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}$

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${\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}$

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Therefore,

The length of the rectangle is 17 inches .

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8 0

1 year ago

Read 2 more answers

On a unit circle, the terminal point of 0 is (sqrt3/2,1/2) What is 0?

san4es73 [151]

Answer:

The measure of $\theta$ is 30°.

Step-by-step explanation:

In the statement, the angle has been misrepresented. The corrected statement is described below:

"On a unit circle, the terminal point of $\theta$ is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$ . What is $\theta$ ?"

The measure of the angle ( $\theta$ ), in radians, is in standard form, that is, it is done with respect to the +x semiaxis. The measure of the angle whose terminal point is of the form $(x,y)$ is determined by the following inverse trigonometric function: