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Nadusha1986 [10]

2 years ago

5

A 16-ounce of shampoo cost $4.69. What is the price per ounce, rounded to the nearest cent

1 answer:

elixir [45]2 years ago

6 0

Answer:

A, 0.29

Step-by-step explanation:

Since 16 ounces cost 4.69, 1 ounce costs 4.69 divided by 16. This is roughly 0.29, which would then be A.

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Please solve the system of equations below thank you<br> explain if u can

ivanzaharov [21]

First let’s try to cancel out the x
5x + -5x = 0
Add the y and the numbers together
-3y + -2y = -5y
26 + -16 = 10
-5y = 10
y = -2
Use y=-2 in one of the equations
-3(-2) + 5x = 26
6 + 5x = 26
5x = 20
X = 4

So
Y= -2
X= 4

https://mathsmadeeasy.co.uk/gcse-maths-revision/simultaneous-equations-gcse-maths-revision/

Check this link out for more help!

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2 years ago

Which equation describes the graph?<br><br> Y=-|x+2|<br> Y=-|x|+2<br> Y=-|x-2|<br> Y=-|x|-2

Lera25 [3.4K]

Answer:

y = -|x|+2

Step-by-step explanation:

The graph represents y = |x| where the given graph opens upside down so it's negative and +2 translation towards the top

Answered by GAUTHMATH

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2 years ago

There are four erasers but only three pencils write the ratio of erasers to pencils

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4:3
4 to 3
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3 years ago

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Read 2 more answers

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