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2 years ago

What is the value of x in the equation 58 = 10x − 2?

Mathematics

2 answers:

solong [7]2 years ago

7 0

Answer:

Step-by-step explanation:

10x6=60-2=58

horrorfan [7]2 years ago

3 0

Answer:

x = 6

Step-by-step explanation:

$58 = 10x - 2 \\58 + 2 = 10x \\60 = 10x \\6 = x$

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-10(4-10x)+5(1-10x)

Ugo [173]

The answer is 50x-35.hope this helps.if it does please mark brainliest

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2 years ago

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A gas has a volume of 5.0 L at a pressure of 50kPa. What happens to the volume when then the pressure is increased 125kPa. The t

Olegator [25]

9514 1404 393

Answer:

volume goes to 2.0 L

Step-by-step explanation:

At constant temperature, volume and pressure are inversely related. Increasing the pressure by a factor of 125/50 = 2.5 will cause the volume to be reduced by that factor.

volume = 5.0L / 2.5

volume = 2.0 L

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2 years ago

If something is $380, then the sellers mark up the price by 84%, what is the final sale price??

nadya68 [22]

699.2 is correct answer

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2 years ago

Find the mean, median, and mode 77, 92, 80, 92. Please explain how you got your answer.

vaieri [72.5K]

Answer:

The mean is 85.25

The medium is 86

The mode is 92

Step-by-step explanation:

The mean is all the numbers added divided by the number of values.

77 + 92 + 80 + 92 divided by 4

341 divided by 4 = 85.25

The median is the middle number

92 + 80 divided by 2

172 divided by 2 = 86

The mode is the number that occurs the most which is 92.

3 0

2 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

2 years ago