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3 years ago

8.50+15.75x=7.50+15.75x

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

4 0

Answer:

No solution

Step-by-step explanation:

You will get zero

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I need this answer following the order of operation. Please help!

natita [175]

Answer:

The answer to the question provided is 1.

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3 years ago

Given f(x)=2x^3-15x^2+22x+15<br> Find all real zeros

mario62 [17]

Answer:

Given function f(x) = 2x^3+15x^2+22x-15. Leading coefficient is p= 2,and constant q is 15. If p/q is a rational zero ,then p is a factor of ...

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3 years ago

Read 2 more answers

How do i get 51 with the numbers 5, 6, 7, 8? Must use all of the numbers and only can use addition, subtraction, multiplication,

exis [7]

Answer:

(8-5)*(6+7+(8-6)+(7-5))

Step-by-step explanation:

8 0

3 years ago

A survey of 100 high school students provided this

HACTEHA [7]

Answer:

35/100

Step-by-step explanation:

You have to find how many juniors there are, so if you add 13, 20, and 2 you get 35. For the denominator you have to find the total number of students, so just add all of the numbers together to get 100. So there is a 35/100 chance that a randomly selected student is a junior.

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2 years ago

israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

4 years ago