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2 years ago

Backpacks originally cost $24 but if 40% back to school sale is going on determine the sale price of the backpack.

Mathematics

1 answer:

Naya [18.7K]2 years ago

4 0

Answer: The price is now 14.40

Step-by-step explanation: The difference is 9.60

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5^2 + 12^2 = c^2 please show work I dont quite know how to finish the ending part off help.

Pavel [41]

5^2 + 12^2 = c^2
25 + 144 = c^2
169 = c^2
take the square root of both sides
+ - 13 = c

3 0

3 years ago

4. Use the distributive property to simplify : 12(2−6).

olasank [31]

Answer:

-48

Step-by-step explanation:

12 x 2=24

12 x 6= 72

24 - 72 = -48

7 0

3 years ago

Which one is lower 31.54 for 2 hours or 43.96 for 3.5 hours

pychu [463]

The answer would be 43.96 for 3.5 hours because for each hour it's 12.56

7 0

3 years ago

Find the constant of proportional (r) in the equation y = rx

RUDIKE [14]

Answer:

Constant of proportionality,

Step-by-step explanation:

Constant of proportionality states that the constant value of the ratio of two proportional quantities x and y,

it is written in the form of y = kx, where k is the constant of proportionality.

Given the equation: .....[1]

where r is the constant of proportionality.

From the table we consider

x = 14 and y = 1.4

Substitute these given values in [1] to solve for r;

Divide both sides by 14 we get;

therefore, the Constant of proportionality,

4 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