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nadya68 [22]
4 years ago
8

Is the point (1,3) solution to the linear equation 5x-9y=32 explain

Mathematics
1 answer:
Sunny_sXe [5.5K]4 years ago
7 0
X=10 , y=2 so it'd be (10,2)
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Helpppppppppppp plzzzzzzzzzzzzzzzzzzz
ZanzabumX [31]

Answer:

Step-by-step explanation:

\left[\begin{array}{ccc}7&31\\142&19\end{array}\right] = 7(19) - 142(31) = <em>- 4269</em>

\left[\begin{array}{ccc}7&11&21\\55&-8&2\\-16&-1&-9\end{array}\right] = 7(- 8)(- 9) + 11(2)(- 16) + 21(55)(- 1) - 21(- 8)(- 16) - 7(2)(- 1) - 11(55)(- 9) = <em>1768</em>

\left[\begin{array}{ccc}9.07&6.02&2.01\\-30.7&2.5&3.5\\3.55&-1.1&2.35\end{array}\right] = 9.07(2.5)(2.35) + 6.02(3.5)(3.55) + 2.01(- 30.7)(- 1.1) - 2.01(2.5)(3.55) - 9.07(3.5)(- 1.1) - 6.02(- 30.7)(2.35) = <em>647.3561</em>

4 0
3 years ago
when graphed on a coordinate plane, Bumby avenue can be represented by the equation y=-4x-7. Primrose Avenue can be represented
erastova [34]

Answer:

Bumby Avenue is represented by the line:

y = -4*x - 7.

Primrose Avenue can be represented with:

8x + 2y = 17.

We may want to find the point where both avenues intersect.

Let's isolate y in this second equation:

2y = 17 - 8x

y = 17/2 - 8x/2 = -4*x + 17/2

Now we can see that the equation for Primrose Avenue has the same slope than the equation for Bumby Avenue, and when two linear equations have the same slope, means that the lines are parallel.

Then we can conclude that Primrose Avenue and Bumby Avenue are parallel, so they never meet eachother.

7 0
3 years ago
A hybrid car can travel 45 miles on 1 gallon of gas. Determine the amount of gas needed for a 500 mile trip
Ilya [14]
12 gallons is needed for a 500 mile trip.
8 0
3 years ago
Please help! I've already answered part a, I don't understand what part b is asking.
Luba_88 [7]

Step-by-step explanation:

So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

As you may know, we cannot divide a number by the value of zero. When the denominator is equal to zero, on the graph this will appear as a vertical asymptote, where x approaches the value that makes the denominator zero, but never actually reaches it.

If you look at each denominator, you can set them equal to zero to find the vertical asymptotes

x+1 = 0

x=-1

There should be a vertical asymptote at x=-1, since it would make two of the denominators equal to -1, but let's divide the two fractions first.

Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

f(x) = (\frac{(x+5)(x-4)}{(x+3)(x-2)})-\frac{1}{x-2}

In this simplified version of the fraction, we can technically define f(-1), but in the original version, since it's not defined there is a removable discontinuity at x=-1, meaning there is no vertical asymptote, but the function is still not defined at f(-1), and there will be a hole at that point.

4 0
2 years ago
One side of a square is shown on the coordinate grid. What is the area of this square in square units
pashok25 [27]

Answer:

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4 0
3 years ago
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