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Fudgin [204]
3 years ago
14

A coordinate plane. A line passes through the point (0, 1) and has a positive slope. Which of these points could that line NOT p

ass through? Check all that apply. (12, 3) (–2, –5) (–3, 1) (1, 15) (5, –2)
Mathematics
2 answers:
maw [93]3 years ago
4 0

Answer: (5, -2) (-3, 1)

Step-by-step explanation: (5,-2) can't work because the y-coordinate of the answer is lower than (0, 1), while the x-coordinate is greater than (0, 1). (-3,1) can't work because the two points create a slope of 0, which is neither positive nor negative so it doesn't work. (12,3) works because it is at a higher angle and is to the upper right of the point, creating a positive slope. (-2,-5) works because it's at the lower left of the point, still a positive slope. (1,15) works because it is similar to the point (12,3), it is higher in both the x-value and the y-value of (0,1).

mario62 [17]3 years ago
4 0

Answer:

*The bolded ones are correct:

(12, 3)

(–2, –5)

(–3, 1)

(1, 15)

(5, –2)

Step-by-step explanation:

use the graph to figure this out

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Answer:

It'll be -0.25

Step-by-step explanation:

Notice how it is to the left of zero, anything to the left of zero is negative. With that in mind, we can eliminate A. The half mark between 0 and -1 represents 0.50, and since it goes by quarters, the E is at the first quarter, which is 0.25. Hopefully this helps!

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Write the equation of a line Parallel to the given line and passes through points (-4, -3); (2, 3). Hints find the slope by usin
julia-pushkina [17]

First, we obtain the gradient (slope) of the first parallel line

\text{gradient, m}_1\text{ = }\frac{y_2-y_1}{x_2-x_1}\text{ = }\frac{3-(-3)}{2-(-4)}=\frac{6}{6}\text{ = 1}

Recall that since both lines are parallel, we have that,

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Thus

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Hence, we can find the equation of the parallel line given that it passes through the points (-4, -3)

Using

\begin{gathered} y\text{ = mx + c} \\ \text{where m = m}_2\text{ = 1} \\ \text{and x = -4 ,  y = -3, we have} \\ -3\text{ = 1(-4) + c} \\ -3\text{ = -4 + c} \\ 4\text{ - 3 = c} \\ c\text{ = 1} \\ \text{Thus, the equation of the line is y = (1)x + 1} \\ y\text{ = x + 1} \end{gathered}

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1 year ago
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