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2 years ago

Please help I will mark brainliest!!! (10 points)

Mathematics

2 answers:

Artist 52 [7]2 years ago

8 0

Answer:

a. h = 700

b. j = 6400

c. g = 50

Step-by-step explanation:

Translate each statement into an equation and solve the equation. Remember that to change a percent to a decimal, you divide the percent by 100, which is the same as moving the decimal point two places to the left. Also, a percent of a number means a percent times the number.

a. 20% of h is 140

20% * h = 140

0.2h = 140

Divide both sides by 0.2

h = 700

b. 5% of j is 320

5% * j = 320

0.05j = 320

Divide both sides by 0.05

j = 6400

c. 30% of g is 15

30% * g = 15

0.3g = 15

Divide both sides by 0.3

g = 50

krek1111 [17]2 years ago

7 0

Answer:

A: 700

B: 6400

C: 50

Step-by-step explanation:

A: You can multiply by 5

B: You can just multiply by 20.

C: Every 10% is 5, because 15(30%)/3 is 5. 5 times 10 is 50.

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Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

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$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

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Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

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Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

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Step-by-step explanation:

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