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larisa86 [58]
3 years ago
15

Which of the following is the simplified fraction its equivalent to 0.315. a) 35/999, b) 35/111, c) 105/333, d) 31/99

Mathematics
1 answer:
ad-work [718]3 years ago
8 0

Answer:

B.

Step-by-step explanation:

A. 35/999 = 0.035...

B. 35/111 = 0.315...

C. 105/333 = 0.315...

D. 31/99 = 0.31...

So, B and C are the same. Based on the information given, B is the simplified form of C.

105/333 = 35/111

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Answer:

(x-2)^{2} + (y+3)^{2} = 4

Step-by-step explanation:

Formula

Centre (a,b) \\Radius r(x-a)^{2} + (y-b)^{2} = r^{2}\\\\Centre (2, -3) \\Radius 2(x-(2))^{2} + (y-(-3))^{2} = 2^{2}\\(x-2)^{2} + (y+3)^{2} = 4

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3 years ago
I NEED HELP ASAP !! IF YOU ANSWER WITHOUT ACTUALLY ANSWERING AND ONLY FOR POINTS YOU WILL BE REPORTED!!
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Answer:

D for the first one and the second one

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3 years ago
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There are 40 nickels in a roll. If you have 250 rolls of nickels, how many nickels do you have?pls help me
gayaneshka [121]

Answer:

10,000 nickles

Step-by-step explanation:

Since there are 40 nickles in a roll and you have 250, think that you have 250 sets of 40. This means that this question is a multiplication problem.

Multiply 40 x 250. This would give you 10,000.

Don't forget units.

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PQRS is a parallelogram. PR and QS are diagonals. Match each element to its value. Tiles value of length of value of length of
ss7ja [257]

Since it is a parallelogram cente at point T, then the measure of PT is equal to the measure of TR. And the measure of QT is equal to the measure of TS.

 

PT = TR

a + 4 = 2a

4 = 2a -a

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PT = TR = 8 units

 

QT = TS

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Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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