10. (02.03) Find the solution of 3 to the square root of x+6=-12 , and determine if it is an extraneous solution. (1 point)
x= -3; extraneous
x = -3; not extraneous
x= 10; extraneous
x= 10; not extraneous

x= 10; extraneous

Step-by-step explanation:

Given the expression:

3√x+6 = -12

We are to find the value of x

Divide both sides by 3

3√x+6/3 = -12/3

√x+6 = -4

Square both sides

(√x+6)² = (-4)²

x + 6 = 16

x = 16-6

x = 10

Hence the solution is x =10 and it is extraneous solution

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: