![Give The Factors For The Numerator [n] And Denominator [d] After Reducing.](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/5C5lTisyhsWC41bjhDLB6Q9n0mz5znlz.png)
Answers
Answer:
The factors for the numerator [n] and denominator [d] after reducing will be:
[tex]\frac{x^2-25}{x^2-4x}\div \:\frac{2x^2+2x-40}{x^3-x}=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}[/tex]
Step-by-step explanation:
Given the expression
[tex]\frac{x^2-25}{x^2-4x}\div \frac{2x^2+2x-40}{x^3-x}[/tex]
[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}[/tex]
[tex]=\frac{x^2-25}{x^2-4x}\times \frac{x^3-x}{2x^2+2x-40}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}[/tex]
[tex]=\frac{\left(x^2-25\right)\left(x^3-x\right)}{\left(x^2-4x\right)\left(2x^2+2x-40\right)}[/tex]
[tex]=\frac{\left(x^2-25\right)x\left(x^2-1\right)}{\left(x^2-4x\right)\left(2x^2+2x-40\right)}[/tex]
[tex]\mathrm{Cancel\:the\:common\:factor:}\:x[/tex]
[tex]=\frac{\left(x^2-25\right)\left(x^2-1\right)}{2\left(x-4\right)\left(x^2+x-20\right)}[/tex]
∵ As factor [tex]\left(x^2-25\right)\left(x^2-1\right)=\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)[/tex]
so the expression becomes
[tex]=\frac{\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)\left(x^2+x-20\right)}[/tex]
∵ As factor [tex]2\left(x-4\right)\left(x^2+x-20\right)=2\left(x-4\right)^2\left(x+5\right)[/tex]
so the expression becomes
[tex]=\frac{\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2\left(x+5\right)}[/tex]
[tex]\mathrm{Cancel\:the\:common\:factor:}\:x+5[/tex]
[tex]=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}[/tex]
Therefore, the factors for the numerator [n] and denominator [d] after reducing will be:
[tex]\frac{x^2-25}{x^2-4x}\div \:\frac{2x^2+2x-40}{x^3-x}=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}[/tex]
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Answer:
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Step-by-step explanation:
Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Step-by-step explanation:
You’re welcome :)
Answer:
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Answer:
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Step-by-step explanation:
Answer:
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Answer:
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Step-by-step explanation:
Answer:
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Answer:
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Answer:
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Step-by-step explanation:
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Answer:
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Answer:
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Answer:
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Step-by-step explanation:
See attachment for complete question
Required
Determine the unit rate
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Where 2 is the common factor.
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Step-by-step explanation:
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Answers
9514 1404 393
Answer:
no
Step-by-step explanation:
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Multiplying by 52, we find ...
x = 52(3.5/32) = 5.6875 . . . quarts needed for 52 cupcakes
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<
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Answers
Answer:
D
Step-by-step explanation:
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First integer =
Second integer =
Third integer =
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Answers
Answer:
49, 51, 53
Step-by-step explanation:
Let the first odd integer be n.
Then the consecutive odd integer will be (n+2).
This will be followed by (n+4).
We know that the sum of these three is 153. In other words:
[tex]n+(n+2)+(n+4)=153[/tex]
Solve for n. Combine like terms:
[tex]3n+6=153[/tex]
Subtract 6 from both sides:
[tex]3n=147[/tex]
Divide both sides by 3:
[tex]n=49[/tex]
Therefore, our first odd number is 49.
So, our consecutive numbers must be 51 and 53.
Hence, our sequence is:
49, 51, 53
Notes:
If we get an even number for n or a non-integer value, then there are no three consecutive odd integers that sum to 153.
Answer:
49, 51, 53
Step-by-step explanation:
All odd numbers:
49 + 51 = 100
100 + 53 = 153