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daser333 [38]
4 years ago
5

For homework Brooke has 15 math problem son social study problems and 5 science problems she has 29 problems in all use mental m

ath to determine how many problems she has for social studies tell which property you used
Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
8 0

Answer:

Brooke has 9 problems for social studies.

Step-by-step explanation:

I used addition. I did 15 + 5 = 20. and after that I just knew that you have to add 9 to 20 to get 29, so I figured out that she has 9 social studies problems.

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Answer:

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Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
Help me plsssssssssssssssss
ser-zykov [4K]
THE ANSWER TO X IS 14

1-3x+15
-3x=12/3=4
2-5x-45
+5=50/5=10
3-4+10=14
5 0
3 years ago
Hi everyone, I'm having trouble with this question and I'm not sure how to do it/where to start. Does anyone have a solution to
Andreyy89

This is quite an interesting problem. I am not sure how high you are in math, but I am going to use calculus I techniques to solve it. First, we need to model an equation. Let P be the total profit and x be every time you increase the cost by $10. If you think about it hard enough you come up with the equation

P(x)=(200-5x)(250+10x)

(200-5x) is the amount of plots you will be able to sell, and (250+10x) is the amount you charge for. So, at x =0

P(0)=(200-5(0))(250+10(0))=(200)(250)=$50,000

This is the initial condition where if we sell 200 plots at $250/plot.

So, this equation makes sense.

Now, let's maximize using the first derivative of the function.

Let's get it into an easily differentiable form.

P(x)=(200-5x)(250+10x)=-50x^2+2000x-1250x+50000\\=-50x^2+750x+50000

From here, differentiate the problem.

P'(x)=-100x+750

Now, set it equal to zero and solve for x.

P'(x)=-100x+750=0\\x=7.5

This a critical point of the function. Let's plug back into the original equation to see what it gives us.

P(7.5)=(200-5(7.5))(250+10(7.5))=(162.5)(325)=52,812.50

You cant sell half a plot, so we need to see what happens if we sell 162 plots and 163 plots, and then compare which one gives us more money.

In order to sell 162 plots

200-5x=162\\x=7.6Plug back into P(x) to see the profit

P(7.6)=(200-5(7.6))(250+10(7.6))=(162)(326)=52,812

Now, do the same for 163 plots

200-5x=163\\x=7.4\\P(7.4)=(200-5(7.4))(250+10(7.4))=(163)(324)=52,812

As we can see, they are the same. So, you can charge either $324 or $326 in rent. But, if your teacher is not looking for a logical answer and you can somehow sell half a plot, you can charge $325 in rent for the maximum profit.

6 0
3 years ago
How do I do both of these??
Semenov [28]
Add them up and multyply them. :)

hope this helped
4 0
3 years ago
Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if
Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula, we get

JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}

JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}

JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}

JK=\sqrt{4+9}

JK=\sqrt{13}

Similarly,

KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6

LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}

JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6

Now, perimeter P of ▱JKLM is

P=JK+KL+LM+JM

P=\sqrt{13}+6+\sqrt{13}+6

P=2\sqrt{13}+12

P=2(3.61)+12

P=7.22+12

P=19.22

P\approx 19.2

Therefore, the perimeter P of ▱JKLM is 19.2 units.

3 0
3 years ago
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