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ladessa [460]
3 years ago
14

How to divide fractions

Mathematics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

Step-by-step explanation:

When dividing fractions, I use a method called Keep Change Flip. Let's use the example 1/2 divided by 3/4 as an example. First, you have to keep the first fraction the same. This is what you currently have: 1/2 / 3/4. Next, you have to change the divison sign into a multiplication symbol. Now, this is what it should look like: 1/2 x 3/4. Then, you have to flip the fraction 3/4 into 4/3. Now, you have 1/2 x 4/3. Multiply the fraction. If you don't know how to, multiply numerator by numerator, denominator by denominator. You should get 4/6. However, this isn't your final answer. You can simplify this by dividing by 2 to get 2/3. So, your answer is 2/3. Hope this helped you and let me know if you need more help! :D

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What types of nouns should be used in effective process writing?

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Use the Distributive Property to solve the equation. 3x-5(x-4)=-9+5x+15
Norma-Jean [14]

Answer:

I keep getting no solution

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

5(x−4)=−9+5x+15

(5)(x)+(5)(−4)=−9+5x+15(Distribute)

5x+−20=−9+5x+15

5x−20=(5x)+(−9+15)(Combine Like Terms)

5x−20=5x+6

5x−20=5x+6

Step 2: Subtract 5x from both sides.

5x−20−5x=5x+6−5x

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Step 3: Add 20 to both sides.

−20+20=6+20

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3 years ago
True or False. If False explain your reason
faust18 [17]
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23 is 12% of what number
Anon25 [30]
\frac { 12 }{ 100 } \cdot x=23\\ \\ \Rightarrow \quad \frac { 3 }{ 25 } \cdot x=23\\ \\ \Rightarrow \quad \frac { 25 }{ 3 } \cdot \frac { 3 }{ 25 } \cdot x=23\cdot \frac { 25 }{ 3 } \\ \\ \Rightarrow \quad x=\frac { \left( 20+3 \right) \left( 25 \right)  }{ 3 } \\ \\ \Rightarrow \quad x=\frac { 500+75 }{ 3 } \\ \\ \Rightarrow \quad x=\frac { 575 }{ 3 }
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EASY BRAINLIEST PLEASE HELP!!
butalik [34]

Answer:

see below

Step-by-step explanation:

<h3>Proposition:</h3>

Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE

<h3>Proof:</h3>

1)The lines AD and BC are parallel and AC their transversal therefore,

\displaystyle  \angle DAC =  \angle ACB \\  \ \qquad [\text{ alternate angles theorem}]

2)The lines AB and DC are parallel and BD their transversal therefore,

\displaystyle  \angle BD C=  \angle ABD \\  \ \qquad [\text{ alternate angles theorem}]

3)now in triangle ∆AEB and ∆CED

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  • \angle EDA=\angle EBC
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therefore,

\displaystyle  \Delta AEB  \cong  \Delta CED

hence,

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Proven

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