How to solve for slope? (pre algebra)

Mathematics

1 answer:

STALIN [3.7K]3 years ago

4 0

Answer:

Ok! When given points, to find the slope, you would use this equation: y2-y1/x2-x1. Let me demonstrate. In this set to find the slope with the coordinates (10,8) and (14,20), the y2 value is 20, and the y1 value is 8, and the x2 value is 14, and the y1 value is 10. So, your equation would look like this: (20-8)/(14-10), which simplifies to 12/4, or 3! So the slope is three, and that's how you do that when using an equation. OR, you could graph them, but that isn't too reliable so I do not recommend trying it, since you may not create the right slope.

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Aloiza [94]

Answer:

see explanation

Step-by-step explanation:

Given the points (a, c ) and (b, c )

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3 0

3 years ago

Find the value of (-8)(-3)(-2)(4). A.17 B.-17 C.192 D.-192 helppp

julsineya [31]

Note: two negatives multiplied together = one positive answer
one negative & one positive multiplied together = one negative answer
Multiply -8 and -3 together

-8(-3) = 24
24(-2) = -48
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D) -192 is your answer

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3 years ago

Read 2 more answers

Write the given trinomial if possible as a square of a binomial or as an expression opposite to a square of a binomial: 15ab-9a^

MrMuchimi

Answer:

$- \bigg(3a - \frac{5}{2} b) \bigg)^{2}$

Step-by-step explanation:

$15ab-9a^2-6 \frac{1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{6 \times 4 + 1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{24+ 1}{4} b^2 \\ \\ = 15ab-(3a)^2-\frac{25}{4} b^2 \\ \\ = 15ab- (3a)^2- \bigg(\frac{5}{2} b \bigg)^2 \\ \\ = - \{ - 15ab + (3a)^2 + \bigg(\frac{5}{2} b \bigg)^2 \} \\ \\ = - \{ (3a)^2 + \bigg(\frac{5}{2} b \bigg)^2 - 15ab \} \\ \\ = - \bigg(3a - \frac{5}{2} b \bigg)^{2}$