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MrRissso [65]
3 years ago
14

Scotty had $2 on his meal card at college.He knew that he could overspend and pay the bill at a later times.He bought a candy ba

r for $1.50 and a bag of chips for $1.75.How much money does he owe the school
Mathematics
2 answers:
Paul [167]3 years ago
6 0
2.00 - 1.50 = 0.50
0.50 - 1.75 = -1.25
Solution: he owns $1.25
makkiz [27]3 years ago
3 0

Answer:

Scotty owes the school $1.25.

Step-by-step explanation:

First we add how much he spent:

$1.50 + $1.75 = $3.25

Now we subtract how much money he has on his card.

$3.25 -$ 2.00 = $1.25

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The distance from earth to the moon is about 22^4 miles. the distance from earth to neptune is22^7 miles. which distance is the
juin [17]
22^4=22x22x22x22
22^7=22x22x22x22x22x22x22
This means that 22^7 is greater and by the amount of 22^3
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3 years ago
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8 0
2 years ago
Read 2 more answers
Please help me out plz
9966 [12]
Point slope formula is y= mx+b. To find the slope, or m, you need to find the "rise over run". rise = y coordinates, run = x coordinates. And the slope equation is y1-y2/x1-x2. So let's say the first point is (x1,y1) and the second is (x2,y2). that would be 35-(-31)/5-(-6)= 66/11 or 6/1 aka up six, across one. That is your slope. So far you have y=6x+b, next plug (5,35) into that equation and solve for b (aka the y intercept). So: 35=6(5)+b. 35-30=b, b=5. So your final equation is y=6x+5.
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 &#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
Please help fast will give brainliest
PolarNik [594]
It’s going to be 19/3>37>6.08171>6.012
6 0
3 years ago
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