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krek1111 [17]
3 years ago
14

Find the solution to the system of equations graphed here:

Mathematics
1 answer:
topjm [15]3 years ago
6 0

Answer: (1,3)

All you're doing is looking for where the two lines cross, or the point of intersection. What you can do is draw a vertical line through this intersection point to see that the vertical line lands on x = 1 on the x axis. At the same time, draw a horizontal line to the y axis and it will get to y = 3.

So together x = 1 and y = 3 pair up to get (x,y) = (1,3)

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3y + 5z + 8y - 3z<br><br> A: 11y + 5z<br> B: 14y + 2z<br> C: 14y + 8z<br> D: 16yz
xxMikexx [17]
Combine like terms:
<span>3y + 5z + 8y - 3z = (3y + 8y) + (5z - 3z) = 11y + 2z

So you would choose A confidently
</span>
6 0
3 years ago
Astronomers measure large distance in light years one light year is the distance that light can travel in one year, or approxima
crimeas [40]

Answer:

78,792,000,000,000 miles

Step-by-step explanation:

(5.88 * 10^12) * 13.4 = 78,792,000,000,000

7 0
3 years ago
Read 2 more answers
12. The sum of three consecutive integers is 12 more than twice the largest integer. Which of the following
baherus [9]

Answer:

OPTION 4

Step-by-step explanation:

Let the three consecutive numbers be n, (n + 1) and (n + 2).

Note that the largest number of the three is (n + 2) and the smallest is n.

So, according to the given data, the sum of all the three integers is equal to 1 more than twice more than the largest number.

Writing it mathematically, we have:

n + (n + 1) + (n + 2) = 12 + 2(n + 2)

This is OPTION (4) and is the answer,

5 0
4 years ago
Find the product. (y3)^2 · y^7
Sedbober [7]
Hey there! :D

When powers are outside the parenthesis, you multiply them. When they are not, you add them.

(y^3)^2 * y^7

3*2= 6

y^6*y^7

Add the powers. 

y^13= the simplest form

I hope this helps!
~kaikers
6 0
3 years ago
Read 2 more answers
Write this fraction in index form
expeople1 [14]

Answer:

7^{-1/5}

Step-by-step explanation:

\frac{1}{\sqrt[5]{7}}=\frac{1}{7^{1/5}}=7^{-1/5}

6 0
1 year ago
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