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GalinKa [24]
3 years ago
8

Find the values of x1 and x2 where the following two constraints intersect.

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer: x1 = 251/26, x2 = -111/26

Step-by-step explanation:

Hi!

As you can see in the figure, the point you are looking for is the intersection of two lines.

The intersection point is found solving this system of linear equations (the point must satisfy both equations):

9x_1 +7x_2=57\\4x_1 + 6x_2 = 13

You can solve it, for example, by the method of substitution:

\text{solve for x1 in the first equation:}\\x_1 = \frac{1}{9}(57 - 7x_2)

Then plug x1 into equation 2, and solve for x2:

\frac{4}{9}(57-7x_2) + 6x_2 = 13\\\text{doing the algebra you get:}\\x_2 = \frac{-111}{26}

Then you use the value of x2 to get x1:

x_1 = \frac{1}{9}(57 - 7x_2)= \frac{1}{9}(57 + 7*\frac{111}{26}) = 251/26\\

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m_a_m_a [10]
C-7=12
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4 0
3 years ago
What is the smallest integer that rounds to 300 when rounded to the nearest hundred?
aleksley [76]

Answer:

The answer is 250.

Step-by-step explanation:

The smallest integer that can be round off to 300 is 250.

e.g

250

≈ 300

249

≈ 200

7 0
3 years ago
Okay so- I picked the yellow one and I don’t know if I’m correct?
Maksim231197 [3]

Answer:

I answered it for u before :D

4 0
3 years ago
I need help on #10
Ede4ka [16]
D = sq.root of((x2 - x1)^2 + (y2 - y1)^2)
sq.root of 13 = sq.root of((x - (-2))^2 + (1 - 3)^2)
= sq.root of((x + 2)^2 + (-2)^2)
= sq.root of (x^2 + 4x + 4 + 4)
= sq.root of (x^2 + 4x + 8)

Now if we square both sides:
x^2 + 4x + 8 = 13
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or x = 1
6 0
3 years ago
Write the slipe-intercept form of the equation of each line given the slope and y-intercept.
Eddi Din [679]

Answer:

1. -1/2

2. 3/-2

3. 1/2/2

4. 7/2

Step-by-step explanation:

Hope it helped

4 0
3 years ago
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