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Anit [1.1K]
4 years ago
10

20 POINTS!!!!! (Voting Brainiest) Two Math Problems!!

Mathematics
2 answers:
borishaifa [10]4 years ago
8 0
Anwsers:
first one: Y = 14

second one, (4,-2)
Murljashka [212]4 years ago
7 0
Y= 14 

(4,-2)

hope that helps

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2/3 (x + 7) = 10 A) 22 B) 8 C)- 1 3 D 41 3 help me quick plz
otez555 [7]

Answer:

D

Step-by-step explanation:

D BECAUSE IT IS D

4 0
3 years ago
Why must s-phase occur before mitosis?<br> ...?
stich3 [128]
Here is the answer of the given question above.
S phase happens<span> after G1 </span>phase<span> and this is a crucial period because this is when DNA is copied. G2 phase then happens, which is prior to mitosis. Once this is complete, this then mitosis happens wherein the cells </span><span>divides the contents of the nucleus between two daughter cells. Hope this answer helps.</span>
6 0
3 years ago
What is 7/8 - 6/7 can anybody tell me
Mars2501 [29]

Answer:

0.01785714285

Step-by-step explanation:

You can round it to 0.18 or something but yeah

4 0
3 years ago
Read 2 more answers
52,259 divided by 215 with remainder
Temka [501]

Step-by-step explanation:

Hope it will help you...

4 0
2 years ago
The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c
Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle.  We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}

\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01

3 0
4 years ago
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