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konstantin123 [22]

2 years ago

5

- x3 + 4x2 – 3x + 12 when x = -2

2 answers:

drek231 [11]2 years ago

8 0

It would be 8 because

-(-2)*3+4*-2*2-3*(-2)+12=8

Hope this helps

If I’m wrong then I’m so sorry

dmitriy555 [2]2 years ago

6 0

Answer: 42

Explanation:

Image below

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3x over 2x-6 plus 9 over 6-2x

Mars2501 [29]

Answer: 3/2
Explanation:
3x/2x-6 + 9/6-2x
= 3x/2x-6 + 9/-(2x+6)
= 3x/2x-6 + -9/2x-6
= 3x/2x-6 - 9/2x-6
= 3x - 9/2x-6
= 3(x-3)/2(x-3) (cancel out x-3)
= 3/2

7 0

2 years ago

Read 2 more answers

5. Adam spent half of his weekly allowance buying lunch. To earn

Gekata [30.6K]

Answer: $11.60

Explanation: First, you want to do 15.1 - 9.3 to find what half of his weekly allowance is, and that would be 5.8, AKA $5.80. Next, you want to multiply that number by 2, which would be 5.8(2) to get 11.6, or your final answer, $11.60.

3 0

2 years ago

If angle DAB measures 34 degrees, what is the measure of arc DB

ollegr [7]

Arc DB will also measure 34 degrees.

4 0

2 years ago

HELP PLZZ ASAP !! a. The value of x is 6 , what is the value of y?

erastovalidia [21]

Answer:

a). y = 4.5

b) Scale factor = $\frac{3}{4}$

Step-by-step explanation:

a). Since, polygon Q is the scaled copy of polygon P

Both the figures will be similar.

Therefore, their corresponding sides will be proportional.

$\frac{x}{y} =\frac{4}{3}$

If x = 6,

$\frac{6}{y} =\frac{4}{3}$

y = $\frac{3\times 6}{4}$

y = 4.5

b). Since, polygon Q is the scaled copy of polygon P,

Polygon Q will be the image of preimage polygon P.

Scale factor = $\frac{\text{Length of a side of the image}}{\text{Length of a side of the preimage}}$

= $\frac{3}{4}$

6 0

2 years ago

Suppose water is added to a tank at 10 gal / min , but leakes out at the rate of 0.2 gal / min for each gallon in the tank . Wha

Nana76 [90]

Answer:

50 gallon

Step-by-step explanation:

We are given that

$\frac{dV_{in}}{dt}=10 gal/min$

Let V(in gallon) be the present volume of tank.

$\frac{dV_{out}}{dt}=0.2 V gal/min$

We have to find the smallest capacity the tank can have if the process is to continue indefinitely.

According to question

$\frac{dV}{dt}$ =Incoming volume-outgoing volume

$\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})$

$\frac{5dV}{V-50}=-dt$

Integrating on both sides

$\int \frac{5dV}{V-50}=-\int dt$

$5ln(V-50)=-t+ln C$

By using the formula

$\int \frac{dx}{x}=ln x+C$

$ln(V-50)=\frac{1}{5}(-t+ln C)$

$ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C$

$ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}$

$ln(V-50)=-\frac{1}{5}t+ln C'$

Where $ln C'=ln C^{\frac{1}{5}}$

$ln(V-50)-ln C'=-\frac{1}{5} t$

$ln\frac{V-50}{C'}=-\frac{1}{5}$

Using the formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{V-50}{C'}=e^{-\frac{1}{5}t}$

$V-50=C'e^{-\frac{1}{5}t}$

Substitute $t=\infty$

$V-50=0$

$V=50$

Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon

7 0

2 years ago