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3 years ago

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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grigory [225]

Answer

Step-by-step explanation:

One solution :

x = -1/3 = -0.333

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

-5*(3*x+1)+4*x-(10*x+2)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - 5 • (3x + 1)) + 4x) - (10x + 2) = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

-21x - 7 = -7 • (3x + 1)

Equation at the end of step 3 :

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Step 4 :

Equations which are never true :

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3 years ago

Read 2 more answers

A shirt that normally costs 30 dollers s on sale for 21.75 what percent of the regular price is the sale price

ivanzaharov [21]

72.5%

You can get this by dividing the new price by the original price and then multiplying by 100.

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3 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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Vitek1552 [10]

Answer:

this is not a right triangle

Step-by-step explanation:

^ = squared

The formula forthe Pythagorean theorem is a^+ b^ = c^

If this is a right triangle this must be true

6^ + 8^ = 14^

36 + 64 = 196

100 is not equal to 196

So this is not a right triangle

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3 years ago