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Bess [88]
3 years ago
9

How many numbers between 100 and 200 can be expressed as powers of 2? Show the steps you followed. Which power of 2 is closest t

o 1000?Show your work For middle school
Mathematics
1 answer:
uranmaximum [27]3 years ago
3 0

Answer:

soo the answer is

2,4,8,16,32,64 128?

Step-by-step explanation:

Square the numbers starting with 10 and ending when you get an answer over 200.

b) Keep squaring numbers until you get to one that equals close to, but less than, 1000.

Hint: The square root of 1,000 = about 31.62.

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antiseptic1488 [7]
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Find the slope of the line that passes through (4, 18) and (7, 10).
hichkok12 [17]

The right answer is -8/3

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3 years ago
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Find the length od ypx. Leave your answer in terms of pi
Liono4ka [1.6K]

your answer is

b = 15 pi

length of ypx = circumfrance of circle - 1/4 of circumfrance of circle

= 2pi r - 2pi r/4 = 6 pi r / 4 = 3 pi r / 2

= 3 pi × 10/2 = 15 pi

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2 years ago
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Draw a line say ab take point c out of it through c draw a line parellel to ab using compass and ruler only
nata0808 [166]

Answer:

give me brainliest

Given a line AB with Point C outside it.

Mark point D on the line AB. Join CD

With D as center, and any radius, draw an arc intersecting AB at E, and CD at F.

With C as center, and same radius as before, draw an arc intersecting CD at G.

Open a compass to length EF

Now,with G as center, and compass opened the same radius as before, draw an arc intersecting the previous arc at H.

Draw a line m passing through C and H.

Thus m is the line parallel to AB, and passing through point C

∴m∣∣AB

Step-by-step explanation:

4 0
3 years ago
Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2
IRISSAK [1]

(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0

Suppose the ODE has a solution of the form F(x,y)=C, with total differential

\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

This ODE is exact if the mixed partial derivatives are equal, i.e.

\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}

We have

\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)

\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)

so the ODE is indeed exact.

Integrating both sides of

\dfrac{\partial F}{\partial x}=(x+y)^2

with respect to x gives

F(x,y)=\dfrac{(x+y)^3}3+g(y)

Differentiating both sides with respect to y gives

\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2

\implies g(y)=-\dfrac{y^3}3-2y+C

\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C

so the general solution to the ODE is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C

Given that y(1)=1, we find

\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13

so that the solution to the IVP is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13

\implies\boxed{(x+y)^3-y^3-6y=1}

5 0
2 years ago
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