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makvit [3.9K]
3 years ago
8

To solve an equation, use the properties of equality. What is the first step to solving 3.7x - 15.9 = 2.97?​

Mathematics
1 answer:
Over [174]3 years ago
8 0

Hello!

3.7x-15.9=3.7

why is 3.7:

3.7x to power 1+1

3.7x to power 0

3.7 * 1

=3.7

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The awnser is 8.50x +9x. i hope this helps
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How to evaluate question number 4
ANTONII [103]

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}

In this case, c=0 and f(x)=\sqrt{5-x}. Then

f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

\dfrac{\sqrt{5-x}-\sqrt5}x\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}=\dfrac{\left(\sqrt{5-x}\right)^2-\left(\sqrt5\right)^2}{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac x{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac1{\sqrt{5-x}+\sqrt5}

This is continuous at x=0, so the limit is the value of the expression at x=0:

\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\lim_{x\to0}\frac1{\sqrt{5-x}+\sqrt5}=-\frac1{2\sqrt5}

5 0
4 years ago
An electron has an atomic mass of 5.4 x 10-4. If a carbon atom has 6 electrons, what is the combined atomic mass of the electron
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<span>6*5.4*10^(-4)
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so option A

to know if you have to increase or decrease the exponent you can write the intermediary steps out:</span>
32.4*10^(-4)=32.4/10^4=<span>(3.24*10)/(10*10^3)
reduce the fraction and you have the above solution
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4 years ago
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Write cp and sp i bought a suit rs 5750.00 and sold it at rs 5890.00?​
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CP = Rs 5750.00

SP = Rs. 5890.00

\rm \blue{\: mark \: brainleist \: }

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Wheel I has diameter 23.4. Wheel ll has diameter 18.0. Round the diameter of each wheel to the nearest inch. About how much fart
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Answer:

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