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Verizon [17]
3 years ago
15

"The symbolic claim, if recognized, would give Russia control of... half the Arctic seabed, [or about] 460,000 square miles." If

the Arctic seabed is shaped like a circle, what is its approximate radius? A. 146 miles B. 732 miles C.541 miles D.383 miles
Mathematics
1 answer:
love history [14]3 years ago
8 0
OOOH!I know the Answer it is A. 146 miles.And double OOOHHHH!!!!! I see your snapchat  icon it looks sooooooooooooo cuteeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Please help, I will mark you brainiest, thank you’
stich3 [128]

Answer:

a

Step-by-step explanation:

6 0
3 years ago
I WILL GIVE BRAINLIEST FOR Correct Answer
VladimirAG [237]
Part A) Find BC, the distance from Tower 2 to the plane, to the nearest foot.

in the triangle ACD
sin16=CD/(7600+BD)--------> CD=sin16*(7600+BD)---------> equation 1 

in the triangle BCD
sin24=CD/BD-----------> CD=sin24*BD---------------> equation 2
equation 1=equation 2
sin16*(7600+BD)=sin24*BD-----> sin16*7600+sin16*BD=sin24*BD
sin24*BD-sin16*BD=sin16*7600----> BD=[sin16*7600]/[sin24-sin16]
BD=15979 ft

in the triangle BCD
cos24=BD/BC---------> BC=BD/cos24-------> 15979/cos24-------> 17491
BC=17491 ft

the answer part 1) BC is 17491 ft

Part 2) Find CD, the height of the plane from the ground, to the nearest foot.

CD=sin24*BD        ( remember equation 2)
BD=15979 ft
CD=sin24*15979 -----------> CD=6499 ft

the answer part 2)  CD is 6499 ft
4 0
3 years ago
Read 2 more answers
Mrs. Cruise prepared gift baskets for her friend. She had 21 mangoes 20 apples and 14 avocados. She places the fruits in each ba
GarryVolchara [31]

Answer:

14

Step-by-step explanation:

I think. sorry if i am wrong

5 0
3 years ago
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
(NEED HELP PLS HELP) i really need help
zalisa [80]

Answer:

the answer I believe is c

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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