All categories

4 years ago

PLEASE HELP ASAP!!!

2 answers:

4 0

$\text{Plug in 3 to y and solve:}\\\\4(3)^3-20+9(3)^2+(25 \div5)\\\\4(27)-20+9(3)^2+(25 \div5)\\\\4(27)-20+9(9)+(25 \div5)\\\\4(27)-20+9(9)+5\\\\108-20+81+5\\\\88+86\\\\174\\\\\boxed{\text{A). 174}}$

lana66690 [7]4 years ago

4 0

Answer:

A) 174

Step-by-step explanation:

I'm assuming 4y3 is 4y³ and 9y2 is 9y²

4×3³-20+9×3²+(25÷5)=

4×3³-20+9×3²+5=

4×27-20+9×9+5=

108-20+81+5=

88+81+5=

169+5=

174

A) 174

You might be interested in

Factor 24m-12p+7224m−12p+7224 - 12p + 72 to identify the equivalent expressions.

lord [1]

Answer:

C, A

Step-by-step explanation:

If you would like me to explain, I could but this would be difficult. Trust me.

7 0

3 years ago

Read 2 more answers

How to multiply mixed fractions

morpeh [17]

Answer:

Step-by-step explanation:

Change each number to an improper fraction.

Simplify if possible.

Multiply the numerators and then the denominators.

Put answer in lowest terms.

Check to be sure the answer makes sense.

8 0

3 years ago

Which set of numbers does -5 belong

tester [92]

Is there any more to this question? Because-5 doesn’t belong anywhere

3 0

3 years ago

Write the equation for the line that passes through the points (4, 5) and<br> (6,9). *

N76 [4]

Answer:

y = 2x + 1

Step-by-step explanation:

first find slops

(9-5)/(6-4) = 4/2 = 2 = m

y = mx + b

5 = 2(2) + b

1 = b

y = 2x + 1

5 0

3 years ago

Read 2 more answers

A consumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 8% will need repairs

Sladkaya [172]

Answer:

Probability that a car need to be repaired once = 20% = 0.20

Probability that a car need to be repaired twice = 8% = 0.08

Probability that a car need to be repaired three or more = 2% = 0.02

a) If you own two cars what is the probability that neither will need repair?

Probability that a car need to be repaired once , twice and thrice or more= 0.20+0.08+0.02=0.3

Probability that car need no repair = 1-0.3=0.7

Neither car will need repair= $0.7 \times 0.7=0.49$

b) both will need repair?

Probability both will need repair = $0.3 \times 0.3=0.09$

c)at least one car will need repair

Neither car will need repair= $0.7 \times 0.7=0.49$

Probability that at least one car will need repair= 1-0.49 = 0.51

6 0

3 years ago