(y 3 - 125) ÷ (y - 5)

Answers

Answer 1

Answer: Hello,

We suppose y ≠5
$(y^3-125)/(y-5)=((y-5)(y^2+5y+25))/(y-5)=y^2+5y+25$

Related Questions

HELP!!Write five different integers that have a mean of -10.
Find the LCD that would eliminate the fractions in this equation.4n-1/7=5/7
Audrey wants to ride her bicycle 44 miles this week. She has already ridden 12 miles. If she rides for 5 more days, write and solve an equation which can be used to determine x, the average number of miles she would have to ride each day to meether goal.
On a desk there are 6 sheets of blue paper, 4 sheets of yellow paper, 5 sheets of green paper, and 2 sheets of pink paper. In how many ways can the paper be stacked?
a woman is expecting triplets. what is the probability that there are 2 girls and 1 boy? what is the probability that all 3 babies are girls

1.)What is the only solution of 2x^2 + 8x = x^2 – 16?

Answers

If you would like to solve the equation 2 * x^2 + 8 * x = x^2 - 16, you can calculate this using the following steps:

2 * x^2 + 8 * x = x^2 - 16
2 * x^2 - x^2 + 8 * x + 16 = 0
x^2 + 8 * x + 16 = 0
(x + 4) * (x + 4) = 0
x = - 4

The correct result would be x = - 4.

Answer:

x= -4

Step-by-step explanation:

took the test:)

three bags of potatoes and four cases of corn cost $40. Five bags of potatoes and two cases of corn cost $34. Find the cost of one bag of potatoes and the cost of one case of corn.

Answers

Use variables and set up equations.

$3p + 4c = 40 \n 5p + 2c = 34$

Next, get one variable with equal coefficients.

$3p + 4c = 40 \n 10p + 4c = 68\ (multiplied\ both\ sides\ by\ 2)$

Then get it alone on one side.

$4c = 40-3p \n 4c = 48-10p$

Now we can use substitution. Since $40-3p = 4c$ , substitute $40-3p$ for $4c$ in the other equation and you'll get an equation with just one variable to simplify and solve.

$40-3p = 68-10p \n 40+7p = 68 \n 7p = 28 \n \boxed{p = 4}$

We can then use $p = 4$ in an earlier equation to find $c$ .

$5p + 2c = 34 \n 5*4 + 2c = 34 \n 20 + 2c = 34 \n 2c = 14 \n \boxed{c=7}$

3p + 4c = 40 ⇒ 15p + 20c = 200
5p + 2c = 34 ⇒ 15p +   6c = 102
      14c = 98
   14    14
      c = 7
          5p + 2(7) = 34
      5p + 14 = 34
      - 14 - 14
      5p = 20
   5 5
      p = 4
      (p, c) = (4, 7)

Ben purchased lunch today at the school cafeteria for $2.75. Before today, Ben owed $9.15 on his lunch account. What is the status of his lunch account after today?

Answers

The status of his lunch account after today will be $11.90

What is mean by Addition?

The process of combining two or more numbers is called the Addition.

Given that;

Ben purchased lunch today at the school cafeteria for $2.75.

And, Before today, Ben owed $9.15 on his lunch account.

Now,

Since, Ben purchased lunch today at the school cafeteria for $2.75.

And, Before today, Ben owed $9.15 on his lunch account.

Hence, The total owed money for lunch = $2.75 + $9.15

= $11.90

Thus, The status of his lunch account after today will be $11.90.

Learn more about the addition visit:

brainly.com/question/25421984

#SPJ2

Answer:

-$11.90

Step-by-step explanation:

Add the two numbers and make it negative because it is owed money

Find the missing value 2+?=-13

Answers

Answer:

-15

Step-by-step explanation:

The sum of the 2 numbers (-13) is a negative, meaning the missing number also has to be negative. To find it, you must ask yourself, what minus 2 would equal 13.

Use angle difference identity to find the exact value.1. tan 15 degrees

2. tan -105 degrees

Answers

Hello,

1)
tan (a-b)=(tan a - tan b)/(1+tan a tan b)

15=45-30
tan 45°= (√2/2)/(√2/2)=1
tan 30°=(1/2)/(√3/2)=√3/3
tan (45°-30°)= (1-√3/3)/(1+1*√3/3)=(3-√3)/(3+√3)=(3-√3)²/6=2-√3

tan (-105°)=tan (75°)= tan (60°+15°)=(tan 60+tan15°)/(1-tan 60°*tan 15°)

tan 15°=2-√3
tan 60°=√3°
tan (-105°)=(√3+2-√3)/(1-√3*(2-√3))=2/(1-2√3+3)=1/(2-√3)=2+√3

Solve x2 + 5x + 6 = 0

Answers

Answer:

Use the quadratic formula

=−±2−4√2

x=−b±b2−4ac2ax=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}x=2a−b±b2−4ac

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

2+5+6=0

x2+5x+6=0x^{2}+5x+6=0x2+5x+6=0

a=1a={\color{#c92786}{1}}a=1

b=5b={\color{#e8710a}{5}}b=5

c=6c={\color{#129eaf}{6}}c=6

=−5±52−4⋅1⋅6√2⋅1

Simplify

Separate the equations

Solve

Solution

=−2=−3