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jolli1 [7]
3 years ago
7

Factor 3y^2 - 4y. plzzzzzz

Mathematics
1 answer:
Lisa [10]3 years ago
7 0

Answer:

y(3y - 4)

Step-by-step explanation:

3y^{2} -4y

Both have y.

Factor y out of 3y^{2} -4y

y(3y-4)

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Please help me asap!!
Alika [10]

Answer:

35

Step-by-step explanation:

what I did was add GJ together and imaged it in my head to see if it would fit... I'm not sure tho

3 0
3 years ago
Read 2 more answers
Use the expression 5(6 + 4x) to answer the following: Part A: Describe the two factors in this expression. (4 points) Part B: Ho
Nikolay [14]

Answer:

Part A: The two factors are 6 and 4x. Part B: Question is confusing Part C: The coefficient of the variable is 4.

Step-by-step explanation:

Factors are what gets multiplied to give a product, therefore, 6 and 4x are the products as they are both being multiplied by 5. I don't understand what you are asking for part b. Part C, the coefficient is any number that is multiplied with the variable next to in. For example, in 16x, 16 is the coefficient and x is the variable. In 9y, 9 is the coefficient and y is the variable.

7 0
2 years ago
Solve the equation <br>5h +2 -h = 22​
astraxan [27]

Answer:

h=5

Step-by-step explanation:

5h +2 -h = 22​

4h+2=22

4h=20

h=5

4 0
3 years ago
Read 2 more answers
Suppose that $3000 is placed in an account that pays 16% interest compounded each year. Assume that no withdrawals are made from
Papessa [141]

Answer:

a) $3480

b) $4036.8

Step-by-step explanation:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Suppose that $3000 is placed in an account that pays 16% interest compounded each year.

This means, respectively, that P = 3000, r = 0.16, n = 1

So

A(t) = P(1 + \frac{r}{n})^{nt}

A(t) = 3000(1 + \frac{0.16}{1})^{t}

A(t) = 3000(1.16)^{t}

(a) Find the amount in the account at the end of 1 year.

This is A(1).

A(t) = 3000(1.16)^{t}

A(1) = 3000(1.16)^{1} = 3480

(b) Find the amount in the account at the end of 2 years.

This is A(2).

A(2) = 3000(1.16)^{2} = 4036.8

4 0
3 years ago
Write 0.088 as a fraction in simplest form.
zhenek [66]
11/25

88/1000
Divide each by 8
11. Times 8
125 times 8
7 0
3 years ago
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