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MAVERICK [17]
3 years ago
7

How to solve the problem

Mathematics
1 answer:
Nady [450]3 years ago
4 0
50 - [(6² - 24) + 9√25]
= 50 - [36 - 24 + 9*5]
= 50 - [12 + 45]
= 50 - 57
= -7
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You are throwing a party at an arcade. You recieve 5 tokens for every dollar spent. Create an equation that describes this situa
aksik [14]

Answer: 5D = T

Step-by-step explanation:

A dollar spent will give 5 tokens.

The amount of dollars spent therefore will increase the number of tokens you by fivefold.

Formula is therefore:

5D = T

Testing it.

Assume you spent 5 dollars, how many tokens would you have:

5D = T

5 * 5 = 25 tokens

7 0
3 years ago
What is divide by 2-digit numbers
AlladinOne [14]
What is divide by-2 digit numbers well it's like (2x4)divided by what ever numbers.
8 0
4 years ago
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I need help i dont understand what's it asking me to do ​
arlik [135]

You have to prove that 2 angles are conguent to the other triangles angles; example: ∠A≅∠C by the additon POC (2 of those) and then one side; example: AB≅BA by the reflexive POE (1 of those)

I cant see the triangles but thats how you do them, also im learning this right now too, look up the triangle congruence things like ASA ≅ SSS≅ HL≅ exc. a chart will explain how to prove them using that:

6 0
3 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
Sean has 5.50 in dimes and quarters. he has 8 more quarters than dimes. how many quarters does he have
MA_775_DIABLO [31]
18 quarters 10 dimes
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4 years ago
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