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iris [78.8K]
3 years ago
11

For kids party of africa, it is also a way to get clean water to their communities. How? The merry-go-rounds pump water out of t

he ground. they are called playmup water systems.
People need clean water for drinking, cooking, cleaning, and bathing. Many people around the world do not have clean water. In some parts of Africa, women and girls have to walk miles to get clean water

People are raising money now for more Playpump systems by 2010, they hope to have 4,000 play pumps in Africa
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from the sentence People hoped to have 4,000 PlayPumps in africa by 2010. can u tell me the answers ples

Question 1: (did) what

Question 2: When?

Question 3: Where?
Mathematics
2 answers:
grigory [225]3 years ago
8 0
You should talk about how people could do mission trips and go and build water despenses out their and to do one every month or do a helpful cleanup so that they are clean for the people
satela [25.4K]3 years ago
7 0
Kids in Africa got clean water from water pumps
they want them by 2010 so before then
in Africa
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What’s the answer ?
elena-14-01-66 [18.8K]

Answer:

It is 9x^3.

Step-by-step explanation:

36x^4 and 45x^3.

GCF of 36 and 45 = 9

GCF of x^4 and x^3 = x^3

GCF of 36x^4 and 45x^2 = 9x^3.

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Angelina_Jolie [31]
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2 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

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