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3 years ago

3x + 3y =15 and x = -3y -1

Mathematics

1 answer:

saul85 [17]3 years ago

4 0

Answer:

(x, y) = (8, -3)

Step-by-step explanation:

You can substitute for x in the first equation:

3(-3y -1) +3y = 15

-6y = 18 . . . . . add 3 and simplify

y = -3 . . . . . . . divide by -6

x = -3(-3) -1 = 8 . . . . find x using the second equation

The solution to this system of equations is (x, y) = (8, -3).

You might be interested in

A house is worth $156,000

kherson [118]

Answer:

$18,000

Step-by-step explanation:

Difference in the years is 192000-156000= 36000

To find rate of change, divide the difference by amount of years in difference.

36000/2017-2015

36000/2

Hence, $18000 per year

Hope it helps!

4 0

3 years ago

Simplify the rational expression. State any excluded values.

ANEK [815]

So on the top, you split it up using the difference of squares rule.
You get (x - 6)(x + 6) on the top.

On the bottom, you can undistribute (pull out) -7x.
This gives you -7x(x-6) on the bottom.

Now you can cancel like terms in the numerator and denominator.
You can cancel the (x-6).
You can also move the negative sign from the 7x to the (x + 6)

This leaves you with -(x + 6) over 7x, or (6 - x) over 7x.

To find excluded values, all you need to know is that you can't divide by zero under any circumstances, and you can't have a zero on the top in a rational expression.
The only values for x that would make either of these statements true are if x = 0 (7 × 0 = 0 on the bottom), or if x = 6 (6 - 6 +0 on the top)

So the answer is (6 - x) over 7x, x ≠ 0, 6

7 0

4 years ago

Is c^5 - 49 a difference of squares? <br> a. yes<br> b. no

gogolik [260]

No. Since we cannot factor,

$c^5-49$ .

Hope this helps.

r3t40

4 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPp

BartSMP [9]

I am not sure is you are supposed to solve or rewrite but for solving its this picture, tell me if its rewriting the problem and ill give you that as well!

3 0

2 years ago

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne

Dvinal [7]

Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

$H_0: \bar x =12.7\\H_a: \bar x >12,7$

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = $\frac{6}{\sqrt{1000} } \\=0.1897$

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

5 0

3 years ago