All categories

2 years ago

Any tips on how to do this?? i have no clue how

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

8 0

Answer:

6.7

Step-by-step explanation:

For this questions we have the oppsotie side and need the adjacent

out of SOH, CAH, TOA we will sue TOA

tan(56)=10/x

tan(56)/10=1/x

10/tan(56)=x

x=6.745 which rounds to 6.7

You might be interested in

A class of 30 swim team members took a lifesaving test. Eighteen of the members had an average score of 92 points. The remaining

inysia [295]

Answer:

88.8%

Step-by-step explanation:

1.) 30-18= 12 which is the number of member that had an average of 84

2.)Then multiply 12 times 84 = 1008

3.) You multiply 18 by 92 which equals 1656

4.)Add 1008 and 1656 together and get 2664.

5.)Divide 2664 / 30 and you get the average of 88.8

Hope this helps :)

plz brainly & like

8 0

3 years ago

Read 2 more answers

A ball was tossed up into the air. The height of the ball as a function of the time the ball is in the air in seconds can be mod

Brut [27]

Answer:

12.5

Step-by-step explanation:

Its the answer I just did it :)

3 0

3 years ago

Read 2 more answers

Whats the square root of 61 rounded to the nearest tenth

adelina 88 [10]

I hope this helps you

49 <61 <64

7^2 <61 <8^2

7<square root of 61 <8

7 0

3 years ago

Read 2 more answers

One hundred divided by twice a number is ten

Ludmilka [50]

$\frac{100}{2x} = 10$

Are you trying to find x, or are you only suppose to translate it?

If you are trying to find x, you first multiply 2x on both sides

100 = 10(2x)

Divide 10 on both sides

10 = 2x

Divide 2 on both sides

5 = x

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

3 years ago