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3 years ago

15

You received $100 for Christmas. You want to buy some new shirts. The cost of each

1 answer:

never [62]3 years ago

8 0

Answer:

5 shirts

Step-by-step explanation:

100-25 is 75 and 75 ÷15 is 5

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Ernesto solves the equation below by first squaring both sides of the equation. \sqrt{\dfrac{1}{2}w+8}=-2 2 1 w+8 =−2square

algol [13]

Answer:

<h2>w = -8</h2>

Step-by-step explanation:

Given the equation solved by Ernesto expressed as $\sqrt{\dfrac{1}{2}w+8}=-2$ , the extraneous solution obtained by Ernesto is shown below;

$\sqrt{\dfrac{1}{2}w+8}=-2\\\\square\ both \ sides \ of \ the \ equation\\(\sqrt{\dfrac{1}{2}w+8})^2=(-2)^2\\\\\dfrac{1}{2}w+8 = 4\\\\Subtract \ 8 \ from \ both \ sides\\\\\dfrac{1}{2}w+8 - 8= 4- 8\\\\\dfrac{1}{2}w= -4\\\\multiply \ both \ sides \ by \ 2\\\\\dfrac{1}{2}w*2= -4*2\\\\w = -8$

Hence, the extraneous solution that Ernesto obtained is w = -8

4 0

3 years ago

Unit 3 | Lesson 9

Charra [1.4K]

The formula d = rt gives the distance traveled in time t at rate r. A bicyclist rides for 1.5 hours and travels 25 miles.
d= D. 37.5 miles per hour
r= 25
t=1.5
25*1.5
=37.5

D.
37.5 miles per hour

5 0

3 years ago

Please answer this question

galben [10]

Answer:

c

Step-by-step explanation:

next

5 0

3 years ago

Think about how you can use absolute value notation to express the distance between two points on a coordinate graph. For each p

vagabundo [1.1K]

Answer:

(4, 9) and (-10, -8)

Step-by-step explanation:

8 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

3 years ago