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NeX [460]
3 years ago
6

Paul started work at company b ten years ago at the salary of 30,000 starting year and 2400 for yearly salary increase at the sa

me time sharla started at company c for 36,000 starting year and 2000 for yearly salary increase who earned more during the last year and how much more
Mathematics
1 answer:
Cloud [144]3 years ago
6 0
Sharla, earned 2,000 more than Paul

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What is a million times all the numbers in the world
kozerog [31]
Well, there isn’t really an end for numbers...

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106 is the exponent equivalent to 1 million

So your question would be:
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7 0
2 years ago
Am i correct? calculus
Advocard [28]
Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

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\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
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\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
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\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
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\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
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\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
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\cfrac{d\theta }{dt}=\cfrac{11}{75}

8 0
3 years ago
Find the product.<br><br><br> (−7)(8) =
Inga [223]

Answer:-56

Step-by-step explanation:The product is your answer

4 0
3 years ago
A + 12.5 = 20 <br> Solve the following equation and show work
makkiz [27]
A + 12.5 = 20
- 12.5 -12.5
———————
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The answer is 7.5
5 0
2 years ago
Read 2 more answers
What is the slope of a line perpendicular to the line with equation y = 1/6 (x) – 2
adell [148]
Let gradient of original line = m = 1/6
Gradient of line perpendicular to this  = -1/m = -6

(Gradient = slope)
3 0
3 years ago
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