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1 year ago

Which of the following is the graph of y = |x - 3|

Mathematics

1 answer:

quester [9]1 year ago

8 0

Answer:

y = | x - 3 | is graphed on the attachment below ↓

Step-by-step explanation:

If you ever need an equation graphed, here's a useful site called Desmos that can do that really well and for free.

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For each value of y , determine whether it is a solution to -21=6y-9 . A -5, B 3 , C -2, D 9

Nezavi [6.7K]

Option C

For each value of y, -2 is a solution of -21 = 6y - 9

Solution:

Given, equation is – 21 = 6y – 9

We have to find that whether given set of options can satisfy the above equation or not

Now, let us check one by one option

Option A)

Given option is -5

Let us substitute -5 in given equation

- 21 = 6(-5) – 9

- 21 = -30 – 9

- 21 = - 39

L.H.S ≠ R.H.S ⇒ not a solution

Option B)

Given option is 3

- 21 = 6(3) – 9

- 21 = 18 – 9

- 21 = 9

L.H.S ≠ R.H.S ⇒ not a solution

Option C)

Given option is -2

- 21 = 6(-2) – 9

- 21 = - 12 – 9

- 21 = - 21

L.H.S = R.H.S ⇒ yes a solution

Option D)

- 21 = 6(9) – 9

- 21 = 54 – 9

- 21 = 45

L.H.S ≠ R.H.S ⇒ not a solution

Hence, the solution for the given equation is – 2, so option c is correct

6 0

3 years ago

Solve the equation for y y +5=-(x – 3)

goldenfox [79]

Answer:

y=-x-2

Step-by-step explanation:

y+5=-(x-3)

y+5=-x+3

y=-x+3-5

y=-x-2

6 0

1 year ago

Read 2 more answers

Help me please please

Studentka2010 [4]

Answer:

It's going to be B and E because it's in the hundreds place not in the tens or whole

6 0

2 years ago

Find the differential coefficient of <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

*BRAIN TEASER*

Mademuasel [1]

Answer:

idk but go dtep by step

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers