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Digiron [165]
3 years ago
6

Angela works a basic week of 40 hours and her hourly rate of pay is $12.50. Calculate her weekly wage.

Mathematics
2 answers:
lutik1710 [3]3 years ago
6 0

Answer:

$500

Step-by-step explanation:

blondinia [14]3 years ago
5 0

Answer:$500

Step-by-step explanation:

To calculate the weekly wage, take the hourly rate and multiply it by the number of hours worked. For this question it would be:

$12.50 x 40

=$500

That is your weekly wage

Hope this helped! <3

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What is the value of x in the equation 2x + 3y=36,when y=6​
levacccp [35]

Answer: x=9

Step-by-step explanation: rewrite the given equation with y=6, 2x + 3(6)=36

Now simplify 2x +18=36

Subtract 18 from both sides of the equal

2x=18

Now divide both sides by 2 and your answer is x=9

4 0
3 years ago
I HAVE 110,000 rolls of film, each roll of film weighs 4.70 ounces. How much would it cost me to have them moved at $0.46 per po
Elena-2011 [213]

Answer:

1 798 260.87$

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1 ounce =0.0625 pounds

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3 0
3 years ago
A three-digit number between 100 and 200 is prime. The number, formed by the last two digits of it is also prime. The two last d
wolverine [178]
113  is one number. 131 is another
137, 173

I think that's all of them.

You can check if you google 'list of primes'.

Hold on  i'm not sure about 113 and 131 Is number 1 a prime number?  I dont think it is , So  that leaves us with  only 137 and 173.


3 0
3 years ago
Helpp with this its for a test review and I gotta turn it in quick
lesya [120]

Answer: 0.5

***If you found my answer helpful, please give me the brainliest. :) ***

5 0
3 years ago
Read 2 more answers
Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant be
Leona [35]

In polar coordinates, the region R is the set of points

\left\{(r,\theta)\mid3\le r\le5,\,0\le\theta\le\dfrac\pi2\right\}

and we have x^2+y^2=r^2 and \mathrm dA=r\,\mathrm dr\,\mathrm d\theta. So the integral, converted to polar, is

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_3^5r\sin r^2\,\mathrm dr\,\mathrm d\theta=\frac\pi2\int_3^5r\sin r^2\,\mathrm dr

Substitute s=r^2 to get

=\displaystyle\frac\pi4\int_9^{25}\sin s\,\mathrm ds=\frac{(\cos9-\cos25)\pi}4

5 0
3 years ago
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