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joja [24]
3 years ago
6

Change the fraction into an equivalent fraction with the denominator w2 + w – 20.

Mathematics
1 answer:
ZanzabumX [31]3 years ago
5 0
Answer:
\frac{(w-3)(w-4)}{(w+5)(w-4)}

If we want to expand the brackets, the fraction would be:
\frac{w^2 - 7w + 12}{w^2 + w - 20}

Explanation:
First, we can note that:
w² + w - 20 can be factorized as follows:
w² + w - 20 = (w+5)(w-4)

The given expression already has a (w+5) in the denominator, therefore, all we need to do is multiply the denominator by (w-4)

Since we want the new fraction to be equivalent to the original one, therefore, as we multiply the denominator by (w-4), we will also multiply the numerator by (w-4) to ensure that the value of the fraction is unchanged.

Based on the above, the new fraction would be:
\frac{(w-3)(w-4)}{(w+5)(w-4)}

If we want to expand the brackets, the fraction would be:
\frac{w^2 - 7w + 12}{w^2 + w - 20}

Hope this helps :)
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The pizza that is a better deal is 12" square pan pizza for $3.95.

Step-by-step explanation:

This is because if you get 2 of the 12" square pan pizzas it will be 24" and you will only have to pay $7.90. but if you bought the 24" square pan pizza you would have to pay $14.95 which isnt a fair price.

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What is x equal to ? Link down below.
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50

Step-by-step explanation:

Because of supplementary angles that line has to equal to 180

180-75 = 105

so 105 = (2x + 5)

you can subtract 5 from each side and get

100 = (2x)

then divide each side by 2 and you get

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you can check it by doing (2 x 50 + 5)

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Estimate 1.133+7.684+4.718
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Find the area of the parallelogram with vertices:________.
marin [14]

Answer:

97.98

Step-by-step explanation:

The area of the parallelogram PQR is the magnitude of the cross product of any two adjacent sides. Using PQ and PS as the adjacent sides;

Area of the parallelogram = |PQ×PS|

PQ = Q-P and PS = S-P

Given P(0,0,0), Q(4,-5,3), R(4,-7,1), S(8,-12,4)

PQ = (4,-5,3) - (0,0,0)

PQ = (4,-5,3)

Also, PS = S-P

PS = (8,-12,4)-(0,0,0)

PS = (8,-12,4)

Taking the cross product of both vectors i.e PQ×PS

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PQ×PS = (20-36)i - (16-(-24))j + (-48-40)k

PQ×PS = -16i - 40j -88k

|PQ×PS| = √(-16)²+(-40)²+(-88)²

|PQ×PS| = √256+1600+7744

|PQ×PS| = √9600

|PQ×PS| ≈ 97.98

Hence the area of the parallelogram is 97.98

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3 years ago
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