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Vsevolod [243]
3 years ago
7

20/4+8/3 answer? I don’t get the question

Mathematics
2 answers:
Neko [114]3 years ago
8 0

Answer:

To Acquire

➠<em><u>T</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>Value</u></em><em><u> </u></em><em><u>of </u></em><em><u>2</u></em><em><u>0</u></em><em><u>/</u></em><em><u>4</u></em><em><u>+</u></em><em><u>8</u></em><em><u>/</u></em><em><u>3</u></em>

<em><u>➠</u></em><em><u>F</u></em><em><u>i</u></em><em><u>r</u></em><em><u>s</u></em><em><u>t</u></em><em><u>l</u></em><em><u>y</u></em><em><u> </u></em><em><u>,</u></em><em><u>we </u></em><em><u>need </u></em><em><u>to </u></em><em><u>find </u></em><em><u>out </u></em><em><u>the </u></em><em><u>LCM </u></em><em><u>of </u></em><em><u>the </u></em><em><u>denominators</u></em>

<em><u>➠</u></em><em><u>1</u></em><em><u>2</u></em>

<em><u>Now,</u></em>

<em><u>➠</u></em><em><u>W</u></em><em><u>e</u></em><em><u>'</u></em><em><u>v</u></em><em><u>e</u></em><em><u> </u></em><em><u>to </u></em><em><u>Multiply</u></em><em><u> </u></em><em><u>the </u></em><em><u>numerator</u></em><em><u> </u></em><em><u>with</u></em><em><u> </u></em><em><u>the </u></em><em><u>number</u></em><em><u> </u></em><em><u>which </u></em><em><u>is </u></em><em><u>divisible</u></em><em><u> </u></em><em><u>by </u></em><em><u>the </u></em><em><u>resulting</u></em><em><u> </u></em><em><u>LCM </u></em><em><u>from </u></em><em><u>the </u></em><em><u>respective</u></em><em><u> </u></em><em><u>denominator</u></em>

<em><u>➠</u></em><em><u>j</u></em><em><u>u</u></em><em><u>s</u></em><em><u>t</u></em><em><u> </u></em><em><u>like </u></em><em><u>,</u></em><em><u>here </u></em><em><u>,</u></em><em><u>1</u></em><em><u>2</u></em><em><u>/</u></em><em><u>4</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u> </u></em><em><u>so </u></em><em><u>we'll </u></em><em><u>Multiply</u></em><em><u> </u></em><em><u>3</u></em><em><u> </u></em><em><u>by </u></em><em><u>the </u></em><em><u>numerator</u></em><em><u> </u></em><em><u>of </u></em><em><u>4</u></em><em><u> </u></em><em><u>i.e </u></em><em><u>2</u></em><em><u>0</u></em><em><u> </u></em><em><u>=</u></em><em><u>></u></em><em><u> </u></em><em><u>2</u></em><em><u>0</u></em><em><u>*</u></em><em><u>3</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em>

<em><u>We'll</u></em><em><u> </u></em><em><u>apply </u></em><em><u>the </u></em><em><u>same </u></em><em><u>thing</u></em><em><u> </u></em><em><u>with</u></em><em><u> </u></em><em><u>the </u></em><em><u>other</u></em><em><u> </u></em><em><u>one</u></em>

<em><u>➠</u></em><em><u>1</u></em><em><u>2</u></em><em><u>/</u></em><em><u>3</u></em><em><u>=</u></em><em><u>4</u></em><em><u> </u></em><em><u>=</u></em><em><u>></u></em><em><u> </u></em><em><u>4</u></em><em><u>*</u></em><em><u>8</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>2</u></em>

<em><u>➠</u></em><em><u>N</u></em><em><u>o</u></em><em><u>w</u></em><em><u> </u></em><em><u>,</u></em><em><u>we </u></em><em><u>will </u></em><em><u>add </u></em><em><u>the </u></em><em><u>resulting</u></em><em><u> </u></em><em><u>numerators </u></em><em><u>and</u></em><em><u> </u></em><em><u>will </u></em><em><u>keep </u></em><em><u>it </u></em><em><u>in </u></em><em><u>a </u></em><em><u>fraction</u></em><em><u> </u></em><em><u>with </u></em><em><u>the </u></em><em><u>denominator </u></em><em><u>=</u></em><em><u> </u></em><em><u>LCM </u></em><em><u>of </u></em><em><u>4</u></em><em><u>&</u></em><em><u>3</u></em><em><u>=</u></em><em><u>1</u></em><em><u>2</u></em>

<em><u>➠</u></em><em><u>6</u></em><em><u>0</u></em><em><u>+</u></em><em><u>3</u></em><em><u>2</u></em><em><u>/</u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em>

<em><u>➠</u></em><em><u>9</u></em><em><u>2</u></em><em><u>/</u></em><em><u>1</u></em><em><u>2</u></em>

➠ <em><u>46/</u></em><em><u>6</u></em>

➠<em><u>2</u></em><em><u>3</u></em><em><u>/</u></em><em><u>3</u></em>

<h2><em><u>➠</u></em><em><u>7</u></em><em><u> </u></em><em><u> </u></em><em><u>⅔</u></em></h2>
Bumek [7]3 years ago
3 0

Answer:

Step-by-step explanation:

if you need to add 2 fractions

20/4+ 8/3

common denominator is 12

60/12+32/12=92/12= 7 8/12  =7 2/3

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In this figure, AB || CD and m/2= 60°.<br> What is mZ6?
liubo4ka [24]

Answer:

The measure of \angle 6\ is\ 60\°.

Step-by-step explanation:

Given,

AB║CD  and

m∠2 = 60°

Solution,

Since, in the figure given in attachment line segment AB is parallel to line segment CD. And t is the transversal line.

The theorem of parallel lines states that,

"If two lines are parallel and cut by a transversal line then the pair of corresponding angles are congruent."

So according to the theorem of parallel lines,∠2 and ∠6 makes a pair of corresponding angle.

And the measure of corresponding angle is equal.

So, m∠2 = m∠6

\therefore m\angle 2 = m\angle 6 = 60\°

Thus the measure of \angle 6\ is\ 60\°.

5 0
3 years ago
Need help with step by step explanation pls
sesenic [268]

Answer:

5 \sqrt{2}

Step-by-step explanation:

\sqrt{50}  -  \sqrt{32}  + 2 \sqrt{8}  \\ 5 \sqrt{2}  - 4 \sqrt{2}  + 4 \sqrt{2}  \\  \sqrt{2}  + 4 \sqrt{2} \\ 5 \sqrt{2}

5 0
3 years ago
Factor x3 – 7x2 – 5x 35 by grouping. what is the resulting expression? (x2 – 7)(x – 5) (x2 – 7)(x 5) (x2 – 5)(x – 7) (x2 5)(x –
mars1129 [50]

The factored expression ofx^3 - 7x^2 -5x +35 is (x^2 - 5)(x - 7)

<h3>What is an expression?</h3>

An expression is an algebraic term used for a mathematical statement that includes addition subtraction multiplication and division

x^3 – 7x^2 – 5x + 35

factor the given equation

x^2(x – 7) – 5(x - 7)

Factor out x - 7

(x^2 - 5)(x - 7)

Therefore, the factored expression is (x^2 - 5)(x - 7)

To learn more about the factored expression visit:

brainly.com/question/723406

7 0
2 years ago
I need help,please help me.​
melamori03 [73]

Answer:

1 5/8

Step-by-step explanation:

subtract

8 0
3 years ago
Read 2 more answers
Y - 3 = -2 ( x + 5 ) in standard form
agasfer [191]

Answer:

2x + y = -7

Step-by-step explanation:

Standard form is Ax + By = C

A represents the coefficient with x

B represents the coefficient with y

C represents the constants

1 - Rewrite to make it easier to work it out

y - 3 = -2 ( x + 5 )

2 - Distribute

y - 3 = -2x - 10

3 - Get the variables on one side and the constants on the other

y - 3 = -2x - 10

+ 2x  + 2x

2x + y - 3 = -10

         + 3   + 3

2x + y = -7

Hope this helps,

7 0
3 years ago
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