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Andre45 [30]
3 years ago
5

Find the y-intercept of the following equation. Simplify your answer. y = -10x -3/7

Mathematics
1 answer:
77julia77 [94]3 years ago
7 0

Answer: (0, -3/7)

Step-by-step explanation:

The Y-intercept would be the value of y when x is at 0, which is when the Y axis is intercepted. -3/7 is the starting position of the function when the the X = 0.

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Help me with this problem?<br> 15 points
kobusy [5.1K]

Answer:

See explanation

Step-by-step explanation:

Since 12 is 3 times greater than 4, you simply need to divide all of the ingredients by 3.

Cactus chunks: 4 1/2 divided by 3 is 3/2 or 1 1/2

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Lizard eggs: 4

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Water: 10/3 or 3 1/3

Hope this helps!

5 0
3 years ago
There are 48 students in a speech contest. Yesterday, 2/3 of them gave their speeches. Today, 3/4 of the remaining students gave
Dmitry [639]

Answer:

24 :))))))))))))))

Step-by-step explanation:

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8 0
3 years ago
Which equation gives the number of 1/4 centimeter that are in 7/8 centimeter?
lesya692 [45]
The answer is B.

remember to use the reciprocal when dividing with fractions
5 0
3 years ago
Read 2 more answers
Help me? answers por favor ?
dem82 [27]
<span>Please, post only one question at a time.

The line passing through (1,4) and (-4,-1) has the following slope:

          -1 - 4          -5
m = ------------ = -------- = +1
          -4 - 1          -5

Find the slope of the line passing thru the other set of points.  Graph both lines.  What do you see?  Do the 2 lines have any particular relationship to one another?

</span>
3 0
3 years ago
Read 2 more answers
Factor the GCF: -6x^4y^5 - 15x^3y^2 + 9x^2y^3
natta225 [31]
<span>-6x^4y^5 - 15x^3y^2 + 9x^2y^3

</span>-6x^4y^5 =  -2, 3, x, x, x, x,  y, y, y, y, y
15x^3y^2 =  -5, 3 x, x, x, y, y
9x^2y^3 =    3, 3 x, x, y, y, y

Each group has a 3 in common and each group has 2 x in common and each group has 2 y in common so the GCF = -3x^2y^2

Divide that out and we get 
\frac{-6x^4y^5 - 15x^3y^2 + 9x^2y^3}{-3x^2y^2} =
2x^2y^3 + 5x - 3y
-3x^2y^2(2x^2y^3 + 5x - 3y)

5 0
3 years ago
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