What is the answer for part a and b

Please help 2=*255/<a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="93a5a6aad3bea6a6a5a6a5a3aa">[email&#

leva [86]

The answer is 45 grace

5 0

2 years ago

triangle ABC was rotated 90 degrees clockwise. Then it underwent a dialtion centered at the origin with a scale factor of 4 what

scoray [572]

Answer: The angles of ΔA'B'C are congruent to the corresponding parts of the original triangle.

Step-by-step explanation:

Given : Triangle ABC was rotated 90 degrees clockwise. Then it underwent a dilation centered at the origin with a scale factor of 4.

A rotation is a rigid transformation that creates congruent images but dilation is not a rigid transformation, it creates similar images but not congruent.

Also, the corresponding angles of similar triangles are congruent.

Therefore, The angles of ΔA'B'C are congruent to the corresponding parts of the original triangle.

4 0

2 years ago

Fwam answered 149 questions correctly on his multiple choice science final that had

aniked [119]

Answer:

he answered 74.5% of questions correctly

6 0

2 years ago

Shirley spelled 12 of her 16 spelling words correctly on the test. What percent grade did shirley get on the test

Black_prince [1.1K]

Answer:

75%

Step-by-step explanation:

percent = part/whole * 100%

percent = 12/16 * 100%

percent = 0.75 * 100%

percent = 75%

Answer: 75%

5 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