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Tatiana [17]
3 years ago
5

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPP! Is 16x^2 - 56x + 49 a perfect square trinomial? How do you know? If it is, factor the expressio

n. Please show your work.
Mathematics
1 answer:
jasenka [17]3 years ago
6 0

No need to fear, thehotdogman93 is here!

The first step is to get rid of those very large numbers. It's going to be very difficult to factor unless we can bring those high numbers down. So lets see if we can factor each term.

So after dividing 49 with every single digit. The only number that divides evenly is 7 and one, and 16 isnt divisible evenly by 7 so that didn't work. Looks like we're gonna have to work with these big numbers.

There is something interesting though about these numbers. 16 and 49 are both perfect squares. 16 is the same as 4^2 and 49 is the same as 7^2. So we can factor the whole trinomial as:

{(4x - 7)}^{2}

If we were to expand this out as:

(4x - 7)(4x - 7)

and multiply it back into the original form. It would match with the expression we started with. The 4's would multiply back into 16x^2 and the 7's would multiply back into 49.

Additionally 4 * -7 is -28, so you can combine two -28x's into the -56x term in the original trinomial.

Thus, the answer is yes you can, and the answer is:

{(4x - 7)}^{2}

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Help me with this... ​
marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \:  - x = 630 - 801 \\ or \:  - x =  - 171 \\ x = 171

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

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Step-by-step explanation:

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Circle has radius r=5 in.

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