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Alika [10]
3 years ago
15

Which expression is equivalent to -3y^2-6y^2+6y

Mathematics
2 answers:
marta [7]3 years ago
5 0
-9y^2+6y is your answer :) have a nice day
AveGali [126]3 years ago
3 0

your answer would be: − 9 y^ 2 + 6 y

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If tge angle at the center is 62° and the radius is 12 cm.. find the length of arc​
insens350 [35]

Answer:

Step-by-step explanation:

r = 12 cm

Theta = 62

Length of arc = \frac{theta}{360}*(2\pi r)

                      =\frac{62}{360}*2*3.14*12

                      = 12.98 cm

6 0
3 years ago
The coordinates of parallelogram PVWZ are P(0, 0), V(-p, q), and W(-p - r, q). Find the coordinates of Z without using any new v
Drupady [299]
For the answer to the question above asking to f<span>ind the coordinates of Z without using any new variables. 
</span>

Vector WZ equals vector VP, which is (p, -q) 
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Will give brainliest to person who answers first!! And 20 points!!!
UkoKoshka [18]
Concept artist tools programmer their you go <span />
3 0
3 years ago
Marcus spent 10 hours doing his homework last week. This week he spent 11 hours doing homework. He says that he spent 110% more
swat32

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Yes he is correct, If it was 100% then it would stay the same 10 hours. But since it's 110%, it is in fact 11.

Step-by-step explanation:

7 0
3 years ago
Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD
Ugo [173]

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = \sqrt{75+25} = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that \frac{1}{2} * BF =  \frac{1}{2} * 5 * \sqrt{75} and BF= 0.5\sqrt{75}. Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is  A= BF *  \frac{AB+CD}{2} = 0.5  \sqrt{75}  * \frac{5+10}{2} = 18.75 \sqrt{3}   cm^{2}

3 0
3 years ago
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