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aleksley [76]
3 years ago
12

1.) What is the solution to the system of equations below? y = 2x + 8 3(-2x + y) = 19

Mathematics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer:

y = 2x + 8

6x + 3y = 19

6x + 3(2x + 8) = 19

6x + 6x + 24 = 19

12x = -5

x = -5/12

y = 2(-5/12) + 8

y = -5/6 + 48/6

y = 43/6

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gulaghasi [49]

464 divide by 16 is 29. So y would be 29


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3 years ago
Choose the slope and y-intercept that correspond with the graph. ​ignore this
krok68 [10]

Answer: The second choice is correct:  slope is 3/4, y-intercept is 1

Step-by-step explanation: The y-intercept is where the line crosses the y-axis.

Look for places where the line intersects the grid at "cross-points" like (-4,-2) (0,1) and (4,4)

The slope is rise/run.

The difference in y-values is the rise. 4-1=3  1-(-2) is 3.

The rise is 3.

The difference in x-values is the run. 4-0=4  0-(-4) = 4

The run is 4

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8 0
3 years ago
Read 2 more answers
Marcie baked several pies. She used 4 cups of flour for the dough. How many pints of flour did she use?
Tcecarenko [31]

Answer:

well if several means 3 then she used 12 pints.

Step-by-step explanation:

3X4=12

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5X4=20 and so on.

3 0
3 years ago
Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).
fomenos

Answer:

\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is 2a=|4--10|.

\implies 2a=|14|

\implies 2a=14

\implies a=7

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is (\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)

We need to use the relation a^2+b^2=c^2 to find b^2.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

c=|6--3|=9

\implies 7^2+b^2=9^2

\implies b^2=9^2-7^2

\implies b^2=81-49

\implies b^2=32

We substitute these values into the standard equation of the hyperbola to obtain:

\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1

\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1

7 0
3 years ago
This is hard can someone help ?
Anastaziya [24]

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Both data sets have the same range: 2–10. Set A's mode of 8 is higher than set B's mode of 6.

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... C. $240

6 0
3 years ago
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