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3 years ago

Solve for e. 9e+4=-5e+14+13e9e+4=−5e+14+13e e=

Mathematics

1 answer:

Kitty [74]3 years ago

5 0

Answer:

e = 10

Step-by-step explanation:

9e+4=-5e+14+13e

Combine like terms

9e+4=14+8e

Subtract 8e from both sides

9e-8e+4=8e-8e +14

e +4 = 14

Subtract 4 from each side

e+4-4 = 14-4

e = 10

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All equivalent fractions of 4/5

stira [4]

Answer:

8/10 or 16/20

Step-by-step explanation:

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2 years ago

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Solve these equation<br>|2x+1|=|x-2|

bagirrra123 [75]

Step-by-step explanation:

2x+1x= -2+1

3x= -1

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$\frac{1}{3}$

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3 years ago

Question 3(Multiple Choice Worth 4 points)

PilotLPTM [1.2K]

Answer:

$\large\boxed{\left(-\dfrac{6}{5},\ -\dfrac{8}{5}\right)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}2x+y=-4&(1)\\y=3x+2&(2)\end{array}\right\qquad\text{substitute (2) to (1)}\\\\2x+(3x+2)=-4\\2x+3x+2=-4\qquad\text{subtract 2 from both sides}\\5x=-6\qquad\text{divide both sides by 5}\\\boxed{x=-\dfrac{6}{5}}\qquad\text{put it to (2)}\\\\y=3\left(-\dfrac{6}{5}\right)+2\\\\y=-\dfrac{18}{5}+\dfrac{10}{5}\\\\\boxed{y=-\dfrac{8}{5}}$

4 0

3 years ago

What is the radius of the circle shown below (and how did you solve it)?

natulia [17]

So you rearrange it as two separate equations to complete the square so
(x-6)^2 -36 + (y-3)^2 - 9 +9 = 0
then move the integers to the other side so its in the right format for the equation of a circle and root it
so radius is 6

8 0

3 years ago

L=<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%20%5C0%7D%20x%5E%7Bsin%20x%7D" id="TexFormula1" title="\lim_{x \to \0} x

zhenek [66]

By applying the exponential and logarithmic functions, we have

$x^{\sin(x)} = \exp \left(\ln \left(x^{\sin(x)}\right)\right)$

Then in the limit,

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \lim_{x\to0} \exp \left(\ln \left(x^{\sin(x)}\right)\right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \ln \left(x^{\sin(x)}\right)\right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \sin(x) \ln(x) \right)$

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\ln(x)}{\csc(x)} \right)$

As x approaches 0 (from the right), both ln(x) and csc(x) approach infinity (ignoring sign). Applying L'Hopitâl's rule gives

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\frac1x}{-\csc(x)\cot(x)} \right) = \exp \left( -\lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} \right)$

Recall that

$\displaystyle \lim_{x\to0} \frac{\sin(x)}{x} = 1$

Then

$\displaystyle \lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} = \lim_{x\to0} \frac{\sin(x)}{\cos(x)} = \lim_{x\to0} \tan(x) = 0$

So, our limit is

$\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp(0) = \boxed{1}$

6 0

3 years ago