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2 years ago

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Determine whether an equation written with one operation can be rewritten as an equivalent equation using a different operation.

Choose True or False for each statement. Multiplication and division are inverse operations. Choose... The equation 3x = 12 can be rewritten as x = 12 ÷ 3. Choose... Addition and subtraction are inverse operations. Choose... The equation x + 1.793 = 2.42 can be rewritten as x – 1.793 = 2.42 – 1.793. Choose...

1 answer:

Kamila [148]2 years ago

5 0

Answer:

Yes, multiplication and division are inverse. yes, for the inverse equation. Yes, addition and subtraction are inverse. yes, for the inverse equation.

Step-by-step explanation:

EXAMPLES.

E.g 21 = 3x division

3x x = 21 inverse

E.g 5+x =8 addition

5=8-x inverse.

Thank you! :)

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How much is 110 hours in days and hours

eimsori [14]

4 days and 14 hours.

Method I used:

1. Do 110÷24 and you should get <span>4.58333333333. Ignore the numbers after the decimal point, and write down 4 as your number of days.
2. Do 4</span>×24 (you should get 96) and subtract that from 110 (you'll get 14). The difference will be the number of hours.

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2 years ago

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

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

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

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

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

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

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

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

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

$\Rightarrow y(1.0)\approx 0.9152$

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3 years ago

5/14 in in simple form

slavikrds [6]

Answer:

Step-by-step explanation:

1. Find the GCD ( or HCF) of numerator and denominator

<u>GCD of 5 and 14 is 1</u>

2. Divide both the numerator and denominator by the GCD

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3 reduced fraction

5/14

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The answer is 22 because in order to divide fractions (in this case we would have 11 as a fraction), you need to multiply one fraction by the reciprocal of the other fraction (the flipped form of a fraction).

To find this, we would do $\frac{11}{1}$ * $\frac{9}{18}$ .

This would come to be $\frac{198}{9}$ , and when simplified by dividing the numerator and denominator by 9, you get 22.

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3 years ago

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ale4655 [162]

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