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3 years ago

Barry Katz earns $1,275 per week. He is married and claims two exemptions. What is Barry’s income tax? Use percentage method.

Mathematics

1 answer:

Nataliya [291]3 years ago

3 0

Answer: I’m not sure but I think the answer is $255

Step-by-step explanation:

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In the figure below, if BE is the perpendicular bisector of AD, what is the value of x?

dedylja [7]

Both sides would be the same, so set them to equal ans solve for x.

3x+4 = 2x +7

Subtract 4 from each side:

3x = 2x +3

Subtract 2x from each side:

x = 3

The answer is X = 3.

4 0

3 years ago

I make $348 a week at my new job. what is my yearly salary?

rosijanka [135]

Answer:

$18,096 a year

Step-by-step explanation:

There are around 52 weeks in a year. You make $348 a week.

52 x 348 = $18,096

3 0

2 years ago

Read 2 more answers

If a = -35, b = 10 cm and c = -5, verify that:

Murljashka [212]

Step-by-step explanation:

(i). a+(b+c) = (a+b)+c

-35+(10-5) = (-35+10)+(-5)

-35+5 = -25-5

-30 = -30

(ii). a×(b+c) = a×b + a×c

-35 × [10+(-5)] = -35×10 + -35×-5

-35 × (10-5) = -350 + 175

-35 × 5 = -350 + 175

-175 = -175

7 0

2 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

The box plot is based on the?<br> mean<br> median<br> mode<br> range

Westkost [7]

PLEASE GIVE BRAINLIST

By reading this ithink the answer is B=median

A boxplot is a standardized way of displaying the dataset based on a five-number summary: the minimum, the maximum, the sample median, and the first and third quartiles. Minimum : the lowest data point excluding any outliers.

hope this helped

3 0

3 years ago