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sergij07 [2.7K]
2 years ago
5

Please help someone, mildy hard

Mathematics
2 answers:
erastova [34]2 years ago
8 0
The answer is 4823.84 I’m pretty sure
Jet001 [13]2 years ago
3 0

Answer:

welp,more piont for me cuz tats confuzizzng

Step-by-step explanation:

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PLEASE HELP 10 POINTS
Ilia_Sergeevich [38]

Answer:

40

Step-by-step explanation:

180 degrees for a triangle. We add 87 and 53 degrees.

That equals 140. Again, back to 180 degrees. We minus 180 degrees with 140 degrees which equals:

40 degrees.

4 0
3 years ago
What is the difference of the fractions? Use the number line and equivalent fractions to help find the answer.
Dmitry_Shevchenko [17]

9514 1404 393

Answer:

  (c)  -3/4

Step-by-step explanation:

Subtracting a positive number moves you to the left on the number line. Subtracting a negative number moves you in the opposite direction, to the right.

Here, we start at -2 1/2 = -5/2, and we move 1 3/4 = 7/4 to the right from there. Each mark on this number line is 1/4 unit, so we move 7 marks. The results is ...

  -2 1/2 -(-1 3/4) = -5/2 +7/4

  = -10/4 +7/4 = -3/4

5 0
2 years ago
Help with this please.....<br> 3=c/-11<br> a.33<br> b.-33<br> c.-14<br> d.14
djverab [1.8K]
The answer is b or c
5 0
2 years ago
PLEASE HELP... VERY URGENT... WILL GIVE BRAINLIEST!!!
lakkis [162]

Answer:

Step-by-step explanation:

a) Pythagorean theorem,

altitude² + base² = hypotenuse²

h²  + (w² - 1)² = (w² + 1)²

                    h² = (w² + 1)² - (w² - 1)²    

{Compare with (a + b)² - (a -b)² = 4ab where a =w² & b = 1}

                   h² = 4*w²*1

                   h² = 4w²

                   h=\sqrt{4w^{2}} = \sqrt{2*2*w*w} \\\\\\h = 2w

b) Area of triangle = \frac{1}{2}b*h

                               =\dfrac{1}{2}*(w^{2}-1) *2w\\\\= (w^{2}-1)*w = w^{2}*w - 1*w\\\\=w^{3}-w

c) w =2

Area of triangle = 3w = 3*2 = 6

7 0
2 years ago
Sin(a+b).cos(a-b) = sinacosa + sinbcosb
Rainbow [258]
Sin (A + B) = sin A cos B + cos A Sin B 
<span>Cos (A - B) = cos A cos B + sin A sin B </span>

<span>=> (SinACosB+ CosASinB) (CosACosB +SinASinB) </span>

<span>=>SinACosACos^2B+Sin^2ACosBSinB+Cos^2A... </span>

<span>=>SinACosA(Cos^2B+Sin^2B) +SinBCosB(Sin^2A+Cos^2A) </span>
<span>we know that Sin^2+Cos^2=1 </span>
<span>=>SinACosA(1)+SinBCosB(1) </span>
<span>=SinACosA+SinBCosB </span>
<span>Proved 

</span>
4 0
3 years ago
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