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anyanavicka [17]
2 years ago
13

3 15. 6x + 10y=0 Find the slipe

Mathematics
2 answers:
Sonja [21]2 years ago
8 0
The answer above me is incorrect, the slope is -3/5x

Here is how i did it:
6x + 10y = 0
-6x. -6x
10y = -6x
——. ——
10. 10
y=-3/5x


Slope intercept form is y=mx+b, and the
original equation was not in that form so 6 is not the slope, it’s is -3/5
rodikova [14]2 years ago
4 0

Answer: y = -3/5x

............................................

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Diane has $1.85 in dimes and nickels. She has a total of 24 coins. How many of each kind does she have?
poizon [28]
<span>Let the nickels be x and the dimes be y
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3 0
3 years ago
Name a point that is the square root of 2 away from(-1,5)
DerKrebs [107]

Answer:

Any point on the circle (x + 1)² + (y - 5)² = 2 with center (-1 , 5) and radius √2

Step-by-step explanation:

circle: (x + 1)² + (y - 5)² = 2

3 0
3 years ago
W=1000M-200 rearrange your make M the subject
Anton [14]

Answer:

M = \frac{w+200}{1000}

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7 0
3 years ago
Find two consecutive even integers
Fantom [35]
You can set up a system of equations with x being the smaller number and y being the larger number. So the first equation would be 2y=5x-14 and the second equation would be x=y-2. You can solve by substitution and substitute y-2 into the x in the first equation to get 2y=5(y-2)-14. You can solve this out and it ends up simplifying to y=8. So you know the larger number is 8. So since the larger number is 8 and they are consecutive even integers, the smaller number must be 6. So the two numbers would be 6 and 8.
8 0
3 years ago
Which table represents a quadratic relationship?
svp [43]

In each case, the x-values are equally-spaced. Thus looking at second differences will tell you if the relation is quadratic. If the second differences are non-zero and constant, then the values have a quadratic relationship.

A. First differences are 2-4 = -2, 1-2 = -1, 0.5-1 = -0.5. Second differences are -1-(-2) = 1, -0.5-(-1) = 0.5. Since 1 ≠ 0.5, this relation is not quadratic. (It is exponential with a base of 1/2.)

B. First differences are 128-135 = -7, 105-128 = -23, 72-105 = -33. Second differences are -23-(-7) = -16, -33-(-23)=-10. Since -16 ≠ -10, this relation is not quadratic. (It is cubic, since 3rd differences are constant at +4.)

C. First differences are -23.2-(-23.4) = 0.2, -23.0-(-23.2) = 0.2, -22.8-(-23.0) = 0.2. Second differences are zero, so this is not a quadratic relation. (It is linear, with a slope of 0.2.)

D. First differences are 56-90 = -34, 26-56 = -30, 0-26 = -26. Second differences are -30-(-34) = 4, -26-(-30) = 4. These are constant (=4), so the relation is quadratic.

The appropriate choice is ...

... D. x -1 0 1 2 3 4

... f(x) 90 56 26 0 -22 -40

4 0
4 years ago
Read 2 more answers
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