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love history [14]
3 years ago
5

20 POINTS !! HELPPP!! Harold worked 27 hours at a rate of $13.25 per hour. Which line in the table will help him calculate his i

ncome tax for this week?
2014 weekly tax table with all amounts in dollars and cents unless noted - first column If your taxable income is (first column Over, second column But not over), second column And your filing status is single (third column Base tax plus fourth column Rate times fifth column Amount over). Line 1 Over 43, But not over 218, Base tax blank, Rate 10 percent, Amount over 43. Line 2 Over 218, But not over 753, Base tax 17.50, Rate 15 percent, Amount over 218. Line 3 Over 753, But not over 1,762, Base tax 97.75, Rate 25 percent, Amount over 753. Line 4 Over 1,762, But not over 3,627, Base tax 350.00, Rate 28 percent, Amount over 1,762. (picture below)



A.


line 1


B.


line 2


C.


line 3


D.


line 4

Mathematics
1 answer:
wariber [46]3 years ago
5 0
<span>Harold worked 27 hours at a rate of $13.25 per hour.

27 * $13.25 = $357.75 so line 2 is the answer.
</span>
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3 years ago
Can anybody help me please
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3 0
3 years ago
Which is larger than 8.15: 55/9 or 57/7 or 41/5 or 65/8​
grin007 [14]
55/9 = 6.11
57/7 = 8.14
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The answer is 41/5 is larger than 8.15
3 0
2 years ago
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Please help me
Furkat [3]
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8 0
2 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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