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3 years ago

13

Maya's standard deduction on her federal income tax return is $8350. If she paid $5980 in state taxes and $2430 in mortgage inte

rest last year, should she use her standard deduction?

2 answers:

Aleonysh [2.5K]3 years ago

5 0

B. No, because it's less than the deduction she would get from itemizing.-APEX

madreJ [45]3 years ago

4 0

Answer: b no because it is less then the deduction she would get from itemizing

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A 20-oz iced tea at a certain restaurant has 90 calories. How many calories are there in a 36-oz iced tea?

kvv77 [185]

Answer:

the answer is 230 calories :)

Step-by-step explanation:

divide 160 by 16 to find out how much 1-oz is then times it by 23 to get the amount of calories in 23-oz. There you go I hope it helps :)

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3 years ago

10) Which two ratios form a proportion? A) 1 : 2 and 4 : 2 B) 1 : 2 and 6 : 3 C) 2 : 1 and 2 : 4 D) 2 : 1 and 6 : 3

Likurg_2 [28]

Answer:

2/1and6/3 because if you do cross multilication they both equal 6

Step-by-step explanation:

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3 years ago

Which complex number does NOT fall within the shaded region in this graph?

avanturin [10]

In order to find which is the answer, we will go step-by-step. Numbers that are not followed by I are drawn on the Real Axis and the other ones are drawn on the Imaginary Axis. You go first on the Real Axis and then on the Imaginary Axis and intersect those 2 points. If a number does not have an imaginary or real part you can assume that is equal to 0. As you can notice the number that is not in the shaded area is <span>2 − I</span>

4 0

3 years ago

Please help! I will not need help with the graph at the bottom just the equation

gladu [14]

isolate y

move y to the right

4y=-2x-8

divide by 4

y = -1/2x -2

solution is

y=-1/2x-2

6 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago