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3 years ago

Why not? If 4 + 6 = 10, does (4 + 6) X 3 = 10 X 3? Why or why not?

Mathematics

1 answer:

IgorLugansk [536]3 years ago

7 0

Answer:

It would be 10*3

Step-by-step explanation:

If we follow the rule of BIDMAS or BODMAS, they both have Brackets to be done first.

By meaning we must do 4+6 as it is in the bracket. Then we multiply our answer by 3.

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what trigonometric expression can be used to find the value of x? replace a and b with the correct values.

Margaret [11]

The trigonometric function that can be used to find the value of x is 12 / tan ( 25 ) . The correct values of a and b will be 12 and 25 resp . .

Given :

A right angle triangle , one angle = 25 ° and side opposite given angle = 12 units .

Trigonometric Formula for Tan θ

tan θ = perpendicular / base , tan θ = ( sin θ / cos θ ) , tan θ = ( 1 / cot θ ) .

where θ = 25 , perpendicular = 12 , base = x .

Substituting values in above first formula ,

tan 25 = 12 / x

Therefore , x = 12 / tan ( 25 ) .

Hence , a = 12 and b = 25 .

To know more on trigonometry :

brainly.com/question/27998034

#SPJ4

4 0

1 year ago

What’s 23-(+25)-(-7)

Tamiku [17]

Answer: It =5

Can I get brainliest please and thank you :)

4 0

2 years ago

Gabriela has dinner at a cafe and the cost of her meal is $45.00. Because of the service,she wants to leave a 15%tip. What is he

Kitty [74]

45+(45*.15)=
45+6.75=51.75
6.75 is the tip, 51.75 is the total bill

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

I donk know what to do

ohaa [14]

What’s the question?

7 0

2 years ago