Ask question

Ask question

All categories

soldi70 [24.7K]

3 years ago

15

Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution. x = 4, solution is ex

traneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous

2 answers:

nydimaria [60]3 years ago

8 0

Answer: The correct option is

(B) solution is x = 4, the solution is not extraneous.

Step-by-step explanation: We are given to solve the following equation :

$\sqrt{2x+1}=3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

And, we are to select the correct option about the solution of the above equation.

From equation (i), we have

$\sqrt{2x+1}=3\\\\\Rightarrow 2x+1=3^2~~~~~~~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow 2x+1=9\\\\\Rightarrow 2x=9-1\\\\\Rightarrow 2x=8\\\\\Rightarrow x=\dfrac{8}{2}\\\\\Rightarrow x=4.$

So, x = 4 is a solution of the given equation.

Now, substituting x = 4 in the left hand side of equation (i), we get

$L.H.S.=\sqrt{2\times4+1}=\sqrt{8+1}=\sqrt{9}=3=R.H.S.$

Since x = 4 satisfy the given equation, so it is not an extraneous solution.

Thus, the required solution is x = 4, the solution is not extraneous.

Option (B) is CORRECT.

Darya [45]3 years ago

5 0

$\text{The domain:}\\2x+1\geq0\to 2x\geq-1\to x\geq-0.5\\\\D:x\geq0.5\to x\in\left\ \textless \ 0.5;\ \infty\right)\\\\\sqrt{2x+1}=3\ \ \ \ |\text{square both sides} \\2x+1=3^2\\2x+1=9\ \ \ |-1\\2x=8\ \ \ \ |:2\\x=4\in D\\\\\text{Answer: x =0, solution is not extraneous}.$

You might be interested in

Alexandra made a tally chart shown below. A 2-column table with 5 rows. Column 1 is labeled Books Read in the Last Month with en

KiRa [710]

Answer:

The Answer is D: "How many books did you read in the last month?"

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Identify the values that create ordered pairs that are solutions to the equation 3x-5y=20

kvasek [131]

The answer to your question is: x=10. y=2

8 0

3 years ago

Read 2 more answers

An equation for the depreciation of a car is given by y = A(1 – r)t , where y = current value of the car, A = original cost, r =

avanturin [10]

Answer:

D

Step-by-step explanation:

We know that the equation for the depreciation of the car is:

$y=A(1-r)^t$

Where y is the current cost, A is the original cost, r is the rate of depreciation, and t is the time, in years.

We are told that the value of the car now is half of what it originally cost. So:

$y=\frac{1}{2}A$

Substitute this for y:

$\frac{1}{2}A=A(1-r)^t$

We also know that the rate of depreciation is 10% or 0.1. Substitute 0.1 for r:

$\frac{1}{2}A=A(1-0.1)^t$

So, let's solve for t. Divide both sides by A:

$\frac{1}{2}=(1-0.1)^t$

Subtract within the parentheses:

$\frac{1}{2}=(0.9)^t$

Take the natural log of both sides:

$\ln(\frac{1}{2})=\ln((0.9)^t})$

Using the properties of logarithms, we can move the t to the front:

$\ln(\frac{1}{2})=t\ln(0.9)$

Divide both sides by ln(0.9):

$t=\frac{\ln(\frac{1}{2})}{\ln(0.9)}$

Use a calculator. So, the car is approximately:

$t\approx6.6\text{ years old}$

Our answer is D.

And we're done!

7 0

3 years ago

Find the derivative. f(x) = e^(x) cosh(x)

zhenek [66]

We are given

$f(x)=e^{x} cosh(x)$

Since, $e^{x}$ and $cosh(x)$ are multiplied

so, we will use product rule of derivative

$\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'$

$f'(x)=\frac{d}{dx}\left(e^x\right)\cosh \left(x\right)+\frac{d}{dx}\left(\cosh \left(x\right)\right)e^x$

we know that

$\frac{d}{dx}\left(e^x\right)=e^x$

and

$\frac{d}{dx}\left(\cosh \left(x\right)\right)=\sinh \left(x\right)$

so, we can plug these values

and we get

$f'(x)=e^x\cosh \left(x\right)+\sinh \left(x\right)e^x$ ...............Answer

7 0

3 years ago

As mush as you can 10 points

kotykmax [81]

24)you add 5a with 5a and then 6 and 4 an and your answer will be -11 -(10a+10)

6 0

3 years ago

Read 2 more answers