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mihalych1998 [28]
3 years ago
7

1. What is the value of the expression 8^–2? a. -16 b. 1/64 c.1/16 d.64

Mathematics
2 answers:
AnnyKZ [126]3 years ago
8 0
Your answer is
B. 1/64
guapka [62]3 years ago
3 0

the answer is 1/64 :)

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Please help I will give you a lot of points and the brainiest.
dimulka [17.4K]

Answer:

The sum is 17,690 nine

Step-by-step explanation:

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3 years ago
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Determine if the following are properties write y or n for each
Paul [167]

Answer:

Answers are below

Step-by-step explanation:

5+1=6       No

78+6=76+8    Yes, commutative property of addition

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52+(-18)=(-52)+18     No

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3 years ago
Brainliest to most helpful! Answer asap!
kap26 [50]

ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA

The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite

Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent

The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle

θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3

α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
8 0
3 years ago
Help me out my smart Brody's
s2008m [1.1K]
That is a perpendicular bisector
4 0
3 years ago
<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28
Nataliya [291]
Assuming you mean

5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t

This ODE is linear in y, and you can already contract the left hand side as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t

Integrating both sides with respect to t yields

5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt
5y\sin t=\dfrac13\sin^3t+C
y=\dfrac1{15}\sin^2t+C\csc t

Given that y\left(\dfrac\pi2\right)=9, we have

9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2
9=\dfrac1{15}+C
C=\dfrac{134}{15}

so that the particular solution over the interval is

y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t
6 0
3 years ago
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