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3 years ago

Answer the following question 11+3p-7=6p+5-3p

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

11+3p-7=6p+5-3p

combine like terms

11-7 +3p = 6p -3p +5

4 +3p = 3p +5

subtract 3p from each side

4 = 5

since 4 does not equal 5

there is no solution

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2 years ago

Which of the following equations has one real solution?

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4x2−12x+9=0

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2 years ago

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State the domain restriction(s) in interval notation of \displaystyle f\left(g\left(x\right)\right)f(g(x)) given: \displaystyle

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Answer:

The interval notation for the domain is $[\frac{23}{3},\infty ]$ .

Step-by-step explanation:

Consider the provided information.

It is given that $\:f\left(x\right)=\sqrt{3x-2},\:\text{ and }\:g\left(x\right)=x-7$

We need to find the value of $f\left(g\left(x\right)\right)$ .

Put the value of g(x) in $f\left(g\left(x\right)\right)$ .

$f\left(g\left(x\right)\right)=f(x-7)$ ....(1)

Now, put x=x-7 in $\:f\left(x\right)=\sqrt{3x-2}$

$\:f\left(x-7\right)= \sqrt{3(x-7)-2}$

$\:f\left(x-7\right)= \sqrt{3x-21-2}$

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From equation 1.

$f\left(g\left(x\right)\right)=\:f\left(x-7\right)= \sqrt{3x-23}$

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Here, the value of $3x-23$ should be greater or equal to 0 as the square root of a negative number is not real.

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The value of x is all real number greater than $\frac{23}{3}$ .

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2 years ago