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Karo-lina-s [1.5K]
3 years ago
9

Decide if 6x + 2 = 2 + 6x has no solution, one solution, or infinitely many solutions

Mathematics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

HI!

Step-by-step explanation:

BYE!

Sorry :(

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(5,-2 is on which quardrant
blondinia [14]

Answer:

Quadrant IV, or the fourth quadrant.

Step-by-step explanation:

Plot your points, first plotting 5, then plotting -2.

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3 years ago
Omg help me thissssss plsssss
igomit [66]

Answer:

3.5

Step-by-step explanation:

If you multiply 3.5 for example by 2 (which is the number of gallons) you get 7 which is the cost (also a proportional relationship).

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-4+ =-1 what is the missing number?
Oksi-84 [34.3K]
T h e. a n s w e r I s t h r e e
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Which side lengths cannot form a triangle?
sineoko [7]

Answer:

answer is option A

A, 4+12>6✓

12+6>4✓

4+6>12×

because sum of two sides lengths ina a triangle should greater than the other side's length

5 0
3 years ago
Read 2 more answers
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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