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3 years ago

Which statements are true the ordered pair (1, 2) and the system of equations? y=−2x+47x−2y=3 Select each correct answer.

Mathematics

1 answer:

netineya [11]3 years ago

4 0

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BRAINLIEST ANSWER!! The product of 3, and a number increased by seven, is -36. Translate the equation, then solve.

EleoNora [17]

Product is multiplication.

Let the number = x

3 *( X+7) = -36

Use distributive property:

3x +21 = -36

Subtract 21 from each side:

3x = -57

Divide both sides by 3:

x = -57 /3

x = -19

Check: 3 * (-19 +7) = 3 * -12 = -36

The number is -19

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3 years ago

Solve for x x+0.15x=3.45

otez555 [7]

Answer: x should equal 3

5 0

3 years ago

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URGENT: Martha is 1 of 5 girls in a backstroke finals of the swim meet. Only the top 2 finishers of the race will receive ribbon

Dima020 [189]

Answer:

50%

Step-by-step explanation:

3 0

3 years ago

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

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

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

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Y=-3x + 4 y= 3x - 2 Solve using Elimination Method

vova2212 [387]

Answer:

x=1

y=1

Step-by-step explanation:

-3x+4=3x-2

-3x+3x+4=3x+3x-2

4=6x-2

4+2=6x-2+2

6=6x

x=1

y=3(1)-2

y=1

3 0

3 years ago