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3 years ago

At the market 8 apples cost $4 how much do 9 apples cost ?

Mathematics

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

$4.5.

Step-by-step explanation:

We have been given that at the market 8 apples cost $4.

Let us find the cost of 1 apple by dividing total cost of apples by total number of apples.

$\text{Cost of 1 apple}=\frac{\$4}{8}$

$\text{Cost of 1 apple}=\$0.5$

To find the cost of 9 apples we will multiply cost of 1 apple by 9.

$\text{Cost of 9 apples}=\$0.5\times 9$

$\text{Cost of 9 apples}=\$4.5$

Therefore, the cost of 9 apples will be $4.5.

castortr0y [4]3 years ago

3 0

If 8 apples cost $ 4, then (4/8) = 1/2 (or 0.50)....so each apple is 0.50 (or 50 cents)

so if u want 9 apples, u will have to pay : 9 * 0.50 = $ 4.50 <=

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Which of the following points could be the initial point of vector v if it has a magnitude of 10 an the terminal point (-1,5)?

Eddi Din [679]

Let the initial point of the vector be (x,y). Then the magnitude of the vector v can be written as:

$|v|= \sqrt{ (-1-x)^{2} + (5-y)^{2} }$

The magnitude of vecor v is given to be 10. So we can write:

$10= \sqrt{ (-1-x)^{2} + (5-y)^{2} } \\ \\ 100=(-1-x)^{2} + (5-y)^{2}$

Now from the given options, we have to check which one satisfies the above equation. That point will be the initial point of the vector.

The point in option d, satisfies the equation.

$100=(-1+7)^{2} +(5+3)^{2} \\ \\ 
100=36+64 \\ \\ 
100=100$

Thus, the answer to this question is option D

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2 years ago

Find the differential coefficient of <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!

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Answer: 73?

Step-by-step explanation:

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