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mihalych1998 [28]
2 years ago
6

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the s

ame solutions as the one shown. 8x + 7y = 39 and 4x – 14y = –68
Mathematics
2 answers:
mylen [45]2 years ago
5 0

Answer:

Using the linear combination method, you can multiply the first equation by 2 and add the equations to get 20x = 10. Dividing both sides by 20, x = 1/2. To solve for y, substitute 1/2 for x in the equation 8x + 7y = 39 to get 4 + 7y = 39. Solving this equation, y = 5. Checking this in the other equation, 4(1/2) – 14(5) = –68 results in 2 – 70 = –68 or –68 = –68. The solution of the system shown is (1/2, 5). The system 8x + 7y = 39 and 20x = 10 is formed by replacing 4x –14y = –68 by a sum of it and a multiple of 8x + 7y = 39. Since 20(1/2) = 10, the system 8x + 7y = 39 and 20x = 10 also has a solution of (1/2, 5).

Step-by-step explanation:

lana66690 [7]2 years ago
4 0
Before you will sum up the following equation, first you must multiply the first equation by two and came up with 16x+14y=78 and then add it to the second equation which is 4x-14y=-68 and the remaining variable is 20x=10 then get the value of X and the value is 1/2 or 0.5. I hope you are satisfied with my answer 
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Subtract 69000 from 60000. Answer becomes 9000. So you add 9000 to 69000 2 times. Your answer becomes 87000.

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The store sells a 9 ounce jar of mustard for $1.53 and a 15 ounce for $2.55 explain whether the cost of the mustards have the sa
astraxan [27]

Answer:

same amount of money

Step-by-step explanation:

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2 years ago
The graph shows the solution to the initial value problem y'(t)=mt, y(t0)= -4
Charra [1.4K]
y'(t)=mt
\\
\\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C

Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.

y=at^2+b
\\
\\3=a\times 2^2-2
\\
\\3=4a-2
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\\5=4a
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\\a= \frac{5}{4} 
\\
\\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C
\\ \\  \frac{m}{2} = \frac{5}{4} 
\\
\\m= \frac{5}{2} 
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\\C=-2
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\\
\\-4= \frac{5}{4} t_0^2-2
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3 0
2 years ago
NEED HELP!
TEA [102]

Answer:

30 minutes

Step-by-step explanation:

20 * 96 / 64 = 30

8 0
2 years ago
The position of an open-water swimmer is shown in the graph. The shortest route to the shoreline is one that is perpendicular to
Elenna [48]

Answer:

y = -\frac{4}{3}x + 9

Step-by-sep explanation:

To create an equation that represents the shortest oath of the swimmer that is perpendicular to the shoreline, we need to find the slope, m, and the y-intercept, b.

The slope of the path would be the negative reciprocal of the slope of the shoreline, since they are perpendicular.

Let's find the slope of the shoreline.

Using two points on the shoreline, (0, 1) and (4, 4),

Slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{4 - 0} = \frac{3}{4}

Therefore the slope, m, of the swimmer's path that is perpendicular to the shoreline would be -⁴/3.

Use the point of the swimmer and the slope of the path to find b.

Substitute m = -⁴/3, (x, y) = (6, 1) in y = mx + b

1 = (-\frac{4}{3})(6) + b

1 = -8 + b

Add 8 to both sides

1 + 8 = b

9 = b

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Now that we know the slope, m = -⁴/3, and the y-intercept, b = 9, plug in their values into y = mx + b.

The equation of the path would be:

y = -\frac{4}{3}x + 9

5 0
2 years ago
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