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3 years ago

A(b-x)=cxmake x the subject of formular

Mathematics

2 answers:

Crank3 years ago

8 0

Answer:

x = $\frac{ab}{c+a}$

Step-by-step explanation:

a(b-x) = cx

ab - ax = cx

ab = cx + ax

ab = x (c + a)

divide both sides by (c + a) for x to stand alone

x = ab/ c + a

Bond [772]3 years ago

3 0

Answer:

×=

$\frac{ab}{?c + a}$

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− 50 67 +1.5−100%

Alex73 [517]

Answer:

<em>Hello, I think the answer is -0.84 Hope That Helps!</em>

8 0

3 years ago

Read 2 more answers

2. Two angle measures are in the ratio 5:9. Write

iris [78.8K]

The two angles are x and 9x/5

Let the two angles we require be x and y.

<h3 /><h3>Ratio of both angles</h3>

We have that the ratio of both angles are x:y

Since both angles are in the ratio 5:9, we have that,

x:y = 5:9

⇒ x/y = 5/9

<h3 /><h3>Value of the other angle</h3>

So, we Make y subject of the formula

Multiplying both sides by y, we have

y × x/y = 5/9 × y

x = 5y/9

Multiplying both sides by 9, we have

9 × x = 5y/9 × 9

9x = 5y

Dividing both sides by 5, we have

9x/5 = 5y/5

y = 9x/5

So, the two angles are x and 9x/5

Learn more about angles here:

brainly.com/question/14362353

3 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 need to practice multiplying fractions

BigorU [14]

Answer:

I am sorry, but that is not a question, if you ask a question about multiplying fractions then I will be happy to answer your question

4 0

3 years ago

Find the center and radius of the circle.<br> (x + 3)2 + (y + 4)2 =9

Tpy6a [65]

The general equation for any circle with its center at point (a, b) is

(x - a)² + (y - b)² = (radius)²

So the circle in the question is centered at (-3, -4), and its radius is 3 .

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3 years ago

A(b-x)=cxmake x the subject of formular​

A(b-x)=cxmake x the subject of formular