Solution set: t>-11/3
A
Use distributive property:
7t-21>4t-32
Then solve for t:
3t>-11
t>-11/3
This is your solution set!
Answer:
d = 14/5
Step-by-step explanation:
The point (-4,2) means that;
At x = -4, y = 2
Now general form of a linear equation is;
Ax + By + C = 0
We are given;
4y = 3x + 6
Rearranging to the form of a linear equation gives;
3x - 4y + 6 = 0
Thus, A = 3, B = -4 and C = 6
Thus, at point (-4,2), distance between them is;
d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)
d = -14/5
We will take the absolute value.
Thus; d = 14/5
It’s like 20 or something
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
![y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}}x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=y%20%3D%20%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7Dx%2B5%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

so we're really looking for the equation of a line whose slope is 3 and passes through (1 , 10)

The answer is D. In an ordered pair, the x-value, which tells you to move left or right, comes first, followed by the y-value, which tells you to move up or down.