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3 years ago

Factor 9b 2 - 4. O (3b + 1)(3b - 4) O (3b + 2)(3b - 2) O (3b - 2)(3b - 2)

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

(3b+2)(3b-2)

Step-by-step explanation: It is the correct answer because we are factoring to the prime number.

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Evaluate the expression:<br> 9 +7(5-2)<br> Using PEDMAS<br><br> yall i suck at math

ANEK [815]

Hello! :)

Let's use PEMDAS.

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

(5-2) = 3 Do parenthesis.

7 × 3 = 21 Multiply with what we got after doing the parenthesis.

9 + 21 = 30 Add with what we had left.

Hope this helps!

ELITEDIPER

5 0

4 years ago

what is an equation to represent for (Andre withdrew spending money from his checking account that initially had $982. What was

Paul [167]

Answer:

The Answer is w - 982 = 922 because it whatever money he took from the 982 lead to 922

Step-by-step explanation:

6 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Ramla bought a sheet of 100 stamps from the post office for $39. what was the price of each stamp?

erica [24]

The price of each stamp was $2.56 each because 100 divided by $39 is $2.56.

Hope that helped:)

5 0

3 years ago

Pls help me with this question ASAP

WINSTONCH [101]

Answer:

Step-by-step explanation:

(3+5)^2/8=(8)^2/8=64/8=8

5 0

3 years ago

Read 2 more answers