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love history [14]
3 years ago
15

Plz help!! I do not know the answer?

Mathematics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

the last one

Step-by-step explanation:

if you look and line it up correctly you can see that they go together

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1/4 + n = 2/3 *using the inverse operation*
Virty [35]

Answer:

Exact Form:

n = 5 /12

Decimal Form:

n= 0.41 6



hope this helps.

sorry if it doesn't

6 0
3 years ago
Read 2 more answers
Find the 8th term of the geometric sequence 10,-40, 160, ...
Mars2501 [29]

Answer:

-163,840

Step-by-step explanation:

Because the pattern is -4 you just multiply it through until you get to the 8th term. ;)

5 0
3 years ago
A rectangle has an area of 72 in². The length and the width of the rectangle are changed by a scale factor of 3.5.
vodka [1.7K]

Answer:

882 in^2

Step-by-step explanation:

4 0
3 years ago
Albert borrowed $12,000 at a
Aleks04 [339]

Answer:

the answer is A

Step-by-step explanation:

8 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
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 
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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