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bearhunter [10]
3 years ago
13

Simplify the algebraic expression:(x^5)^-2​

Mathematics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

1/x^10 should be the answer

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400g of rapberries and 300g of strawberries cost £7.46. 500g of strawberries cost £4.10 work out the cost of 300g of raspberries
MAXImum [283]

Answer: £5.39

Step-by-step explanation:

500g strawberries = 4.10

100 g = 0.82

300g = 2.46

400g raspberries = 7.46 - 2.46 = 5.00

100g = 1.25

300g = 3.75

200g strawberries = 0.82 * 2 = 1.64

3.75 + 1.64 = 5.39

8 0
3 years ago
There are 8 2/3 pounds of walnuts in a container which will be divided equally into containers that hold 1 1/5 pounds this would
igomit [66]

you want 8 2/3 divided by 1 1/5 =

26/3 / 6/5

= 26/3 * 5/6 = 130 / 18

7 and 4/18 containers

4 0
3 years ago
Andreus made 625$ mowing yards. He worked for 5 days and earned the same amount of money each day.How much money did he earn per
Alexandra [31]
Just divide 625$ by 5. The answer is 125$ per day.
5 0
3 years ago
Read 2 more answers
Layla went shopping for camping equipment. After looking at several different stores, she bought a tent for $163.63 and a sleepi
Ghella [55]

Answer:

$297.09

Step-by-step explanation:

To answer this question, just add:

163.63 + 133.46 = 297.09

So, Layla spent $297.09 in all

Hope this helps :)

5 0
3 years ago
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
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