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TEA [102]
3 years ago
10

How would I solve the system 9x +y = 2, -4x -y =-17 when using elimination ​

Mathematics
1 answer:
larisa [96]3 years ago
7 0

Answer: y=29, x=-3

Step-by-step explanation:

eq. 1: 9x+y=2

9 • -3+y=2

-27+y=2

y=29

eq. 2: -4x-y=-17

4x+y=17

5x=-15

x=-3

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Given the line 2x - 3y - 5 = 0, find the slope of a line that is perpendicular to this line. ​
sweet [91]

Answer:

The slope of the line would be -3/2.

Step-by-step explanation:

Let's get this equation into the proper slope form:

2x - 3y - 5 = 0

2x - 3y + 3y - 5 = 0 + 3y

2x - 5 = 3y

(1/3)(2x - 5) = 3y * 1/3

2/3x - 5/3 = y

Since the slope is 2/3, we need to find the negative reciprocal of that number, which is -3/2.

8 0
2 years ago
Read 2 more answers
at the cafeteria, muffins are sold in the following flavors: blueberry, orange, carrot, corn, bran, and choc chip. there is a re
Colt1911 [192]
There are 6 flavors, and each one can be regular or low fat.
So there are 12 different muffins possible.
4 0
4 years ago
Please give answer of 6 number​
Irina18 [472]

Answer:

7,500

Step-by-step explanation:

Let's solve this problem step-by-step. The library had 1,500 books in 2011. The ratio of books in 2011 and in 2012 is 1:2. Therefore, let the number of books in 2012 be x.

We have the following proportion:

\frac{1}{1,500}=\frac{2}{x},\\x=2\cdot 1,500=3,000

Therefore, there were 3,000 books in 2012. The ratio of books in 2012 and in 2013 is 2:5. Let the number of books in 2013 be y.

We have:

\frac{2}{3,000}=\frac{5}{y},\\2y=5\cdot 3,000,\\2y=15,000\\y=\boxed{7,500}

Therefore, there were 7,500 books in 2013.

8 0
3 years ago
Read 2 more answers
Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0
3 years ago
Corey ate a chocolate candy bar that he got for Valentine’s Day. He ate 4/10 of the candy bar on Monday. On Tuesday he ate anoth
VladimirAG [237]
Solutions =2/5×0.3=
3/25
7 0
3 years ago
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