All categories

3 years ago

Eliminating Fractions from Equations

Mathematics

1 answer:

garri49 [273]3 years ago

7 0

Answer:

12 is the best answer

I think

You might be interested in

What is the output, or y-value, when you input x = 1 into the function:

Aleksandr [31]

Answer: $y=-6$

Step-by-step explanation:

By definition, a relation is a function if and only if each input value has one and only one output value.

For this exercise it is important to remember that thw "Input values" are the values of the variable "x"and the "Output values" are the values of the variable"y".

In this case you have the following function provided in the exercise:

$y = x^4 - 2x^3 + x^2 + 4x - 10$

Then, to solve this exercise, you need to follow these steps:

Step 1. You have to substitute $x=1$ into the function given.

Step 2. Finally, you must evaluate in order to find the corresponding output value (or the value of the variable "y")

You get that this is:

$y = (1)^4 - 2(1)^3 + (1)^2 + 4(1) - 10\\\\y=-6$

3 0

3 years ago

Addi places the letters from the word DECEMBER into a bag, a letter will be randomly selected and not replaced. Thnen another le

yuradex [85]

Answer:

Step-by-step explanation:

for c it would be 1\8

for E it would be 3\8

4 0

3 years ago

Read 2 more answers

Enrollment at a school last year was 1220 this year the student population decreased by 10% how many total students are attendin

Umnica [9.8K]

I think 37%......................................................

8 0

3 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

3 years ago

About 18% of the population of a large country is nervous around strangers

sashaice [31]

Answer:

Step-by-step explanation:

4 0

2 years ago