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dybincka [34]
2 years ago
8

Alicia is a nurse.

Mathematics
2 answers:
sattari [20]2 years ago
6 0

Answer: $19.15/hr

Step-by-step explanation:

(1750 * 0.02) + 1750 = 1785 pay increase

1785 * 12 months = 21420 a year

21420 a year divided 52 weeks in a year averages to 411.92 weekly

Divide 411.92 by the 45 hours of work is 19.15 per hour

SOVA2 [1]2 years ago
5 0

Answer:

See below

Step-by-step explanation:

2% increase on 1750=

1750 * 1.02 = 1785

1785 per month gives 1785 * 12 = 21420 per year.

Divide 21420 by 48 weeks = 446.25 per week

Divide 446.25 by 45 hours = 9.92 per hour

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The capacity of a beamer is 0.1 liter. convert this to milliliters
Vladimir [108]

A milliliter is a thousandth of a liter. 1,000 milliliters (mL) are in a liter. So, multiply 0.1 by a thousand to find the answer. The answer is 100.

3 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Write 25/20 as a percentage
kolezko [41]
125% my guys im sorry if its wrong :(
5 0
2 years ago
Read 2 more answers
Please help me with this question I’m so confused
zavuch27 [327]

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 10( \frac{1}{2} ) +  \frac{1}{2} ( - 2) {( \frac{1}{2}) }^{2}  \\ s = 10( \frac{1}{2} ) +  \frac{1}{2} ( - 2)(  \frac{1}{4} ) \\ s = 5 + ( - 1)( \frac{1}{4} ) \\ s = 5 + ( -  \frac{1}{4} ) \\ s = 5 -  \frac{1}{4}  \\ s =  \frac{20}{4}  -  \frac{1}{4}  \\ s =  \frac{19}{4}

Hope it helps

Please give brainliest

4 0
1 year ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%7B3x%7D%5E%7B3%7D%20%20%5Ctimes%204%20%7By%7D%5E%7B5%7D%20%20%5Cdiv%20%20%7B8y%7D%5E%7B2%7
garik1379 [7]
The answer is 12x3y5 divided by 8x15x5y2
6 0
3 years ago
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