Solve the system of equations. -9 - 6y = 15 9x

Solve the system of linear equations by substitution

ziro4ka [17]

$y - x = 0 ...equation 1$
$2x - 5y = 9 \: ...equation \: 2$
working with equation 1
$y = x \: ...equation \: 3$
next subsitue equation 3 in equation 1
$2x - 5(x) = 9$
$2x - 5x = 9$
$- 3x = 9$
divide by -3
$x = - 3$
substitute x= -3 in equation 1
$y - ( - 3) = 0$
$-(-3) = + 3$
$y = - 3$

6 0

3 years ago

Simplify the following math expression: 5+4(6-3)+4/2

lesantik [10]

Answer:

29

you have to solve the brackets first(6-3=3)

Then solve the division (4/2=2)

lastly the addition (5+4=9)

so we have 9×3=27

27+2=29.

3 0

2 years ago

Read 2 more answers

Carl and rose live in the same straight road. From their balconies they can see a flagpole in the distance. The angle that each

Andreyy89

Carl and Rose' balconies make up the base of an isosceles triangle.

Their distances from the flagpole is the same.

From the question, we understand that:

Carl and Rose live on a straight line
The measure of angle from each person's balcony to the flagpole is the same

The above highlights mean that:

The relationship between Carl and Rose' balconies and the flagpole is an isosceles triangle.

Where Carl and Rose' balconies form the base of the isosceles triangle.

Hence, their distances from the flagpole is the same.

Read more about distances at:

brainly.com/question/12961022

6 0

3 years ago

How do you substitute the equation x=y-1

umka2103 [35]

If you need to substitute the equation x = y - 1.
If you have another equation, then you would plug in y - 1 into the x value of the equation you would need to substitute into.

Hope this helps

7 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago

Solve the system of equations. -9 - 6y = 15 9x – 10y = 145