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3 years ago

15

the area of a triangle is 108square feet if the height is 6 less than twice the base then find the base and the height of the tr

iangle

1 answer:

bulgar [2K]3 years ago

6 0

Base = 12

Height = 18

12*18= 216

216/2= 108

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1kg= 2.2 pounds<br> Change 44 pounds in kg

Lyrx [107]

<u><em>Answer:</em></u>
44 pounds = 20 kg

<u><em>Explanation:</em></u>
<u>We are given that:</u>
1 kg = 2.2 pounds

To convert 44 pounds into kg, all we have to do is <u>cross multiplication</u> as follows:
1 kg .................> 2.2 pounds
?? kg .................> 44 pounds

<u>Now, we solve as follows:</u>
4 kg = $\frac{44*1}{2.2} = 20$ pounds

Hope this helps :)

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4 years ago

Which best describes the error in finding the simple interest earned on $500 at 6% for 18 months?

valentinak56 [21]

$540.00 is the interest
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3 years ago

If f(x)=x^2- 4 and g(x)=x^2 plus 2 x,

gregori [183]

Answer:

$\huge\boxed{(f-g)\left(-\dfrac{1}{3}\right)=-3\dfrac{1}{3}}$

Step-by-step explanation:

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3 years ago

The slope of a line is 5 /7. What is the slope of any line that is perpendicular to that line?

Lady bird [3.3K]

Answer:

-7/5

Step-by-step explanation:

The slope of a perpendicular line is the "negative reciprocal" of the slope of the original line. so just switch the numbers and put the top number into a negative.

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4 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

3 years ago

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