Answer:
<em>S</em><em>o</em><em> </em><em>1</em><em>)</em><em> </em><em> </em><em>x</em><em>²</em><em>-</em><em>2</em><em>x</em><em>-</em><em>8</em>
<em>=</em><em> </em><em> </em><em> </em><em> </em><em> </em> x(x-2-8/x)
<em>2</em><em>)</em><em> </em><em> </em><em> </em><em>y</em><em>²</em><em>-</em><em>1</em><em>3</em><em>y</em><em>+</em><em>4</em><em>2</em>
<em> </em><em>=</em><em> </em><em> </em><em> </em>y(y-13+42/y)
<em>3</em><em>)</em><em> </em><em>m</em><em>²</em><em>-</em><em>6</em><em>m</em><em>-</em><em>7</em>
<em> </em><em>=</em><em> </em><em>m</em><em>(</em><em>m</em><em>-</em><em>6</em><em>-</em><em>7</em><em>/</em><em>m</em><em>)</em>
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em>
Give me a minute. I’m trying to solve
The correct answer is: " 
" .
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<u>Step-by-step explanation</u>:
Based on the assumption that the "1" repeats infinitely; in the given value:
" 33.61111111 ...." ;
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Note that the "611" ; after the decimal point; this goes to the "thousandths";
place (is "3 (three) digits long.").
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As such; we rewrite the number as:
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" 
" ;
and we multiply BOTH the "numerator" And the "denominator" by: "1000" :
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→ " 
" ;
to get:
→ " 
" ; → which cannot be reduced any further.
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The correct answer is: " " .
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Hope this is helpful to you!
Wishing you the best!
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Answer:
Step-by-step explanation:
A=(3.14d^2)/4
A=0.785d^2
A=0.785(10^2)
A=78.5 yd^2
Hey there!
When we're adding with different denominators, our goal is to keep the equivalent fraction, but create like denominators.
Let's think of an easier situation. If we have the number 5 and we want an equivalent number, we multiply by one. It's no different with fractions. We want to multiply by some version of one, like 2/2 or 4/4
For example, if we have:
2/8 + 4/6
Our LCM is 24. Therefore, we multiply 2/8 by 3/3:
2/8(3/3) = 6/24
And 4/6 by 4/4:
4/6(4/4) = 16/24
As you can see, we multiplied by versions of 1, so they're still the same fraction.
We have:
16/24 + 6/24 = 22/24 = 11/12
Hope this helps!
