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Reil [10]
3 years ago
12

What is the value of x that will make A parallel to B?

Mathematics
1 answer:
Zanzabum3 years ago
5 0

Answer:

10 = x

Step-by-step explanation:

Since they are parallel, set them equal to each other:

3x + 10 = 4x

-3x - 3x

_________

10 = x

I am joyous to assist you anytime.

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I need help with number 6 pls help!!!!!
Oksi-84 [34.3K]

Answer:

for 6 it would be 144

Step-by-step explanation:

I'm pretty sure you have to mutiply all of them

3 0
2 years ago
You have 2\3 of a pizza left from yesterday. You divide it into 4 equal pieces. What fraction of the pizza is each piece?​
Mrrafil [7]

Step-by-step explanation:

that means we divide 2/3 by 4.

2/3 / 4 = 2/3 / 4/1 = 2/3 × 1/4 = 2/12 = 1/6

so each quarter of the remaining 2/3 pizza is actually 1/6 of the complete pizza.

7 0
2 years ago
Match the equation with the graph -0.4x-0.8y=0.10
laiz [17]
Can't see the graph, but it should look something like this:

7 0
2 years ago
The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model
Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$  in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

                    $=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

                   $=10,000[e^{0.4}-1]$

                    $=10,000[0.49]$

                    =4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

   $=10,000e^{0.02(5)}$

   $=11051$

In 2000,

$P=10,000e^{0.02(10)}$

   =12,214

Therefore, the change in the population between 1995 and 2000 = 1,163.

8 0
3 years ago
-5(x + 1) s 30. What is a possible value of x?<br> help me pls
inn [45]

Answer:

the answer should be x= -7 hope this helps!

4 0
3 years ago
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