The slope of a line perpendicular to the
graph of the equation 5x - 3y = 2 is -3/5.
<h3>How to find the slope of a line?</h3>
given that the equation is 5x - 3y = 2.
now write the equation in standard form y = mx + b
then -3y = 2 - 5x
y = -2/3 + 5x/3
y = 5/3x - 2/3
m 1*m 2 = - 1 is the formula for the slopes from a pair of perpendicular lines. where the slopes of the lines are m 1 and m 2.
Here m1 = 5/3 and m2 = -3/5.
Hence,the slope of a line perpendicular to the
graph of the equation 5x - 3y = 2 is -3/5.
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Answer:
y=2/3x+14
Step-by-step explanation:
The standard form of an equation in slope-intercept form is y=mx+b where m=slope and b=y-intercept.
Given a y-intercept of 14 and a slope of 2/3, we can plug into the variables and get the equation y=2/3x+14
The x-intercept would be when y=0, so plugging in y=0 to the equation gets us:
0=2/3x+14
-14=2/3x
21=x
So the x-intercept is 21
Answer:
B
Step-by-step explanation:
the domain of a graph consists of all the input values shown on the x-axis.
so because the only parts shown on the x axis are 6 and -6 that is the answer!
hope this helps!
A. Area of ABCD - Area of DGA = Area of DEFG
s^2 - 1/2bh = s^2
(5)^2 - 1/2(4)(3) = (3)^2
25 - 1/2(12) = 9
25 - 24 = 9
1 not equal to 9
B. Area of ABCD - Area of GHIA = Area of DGA
s^2 - s^2 = 1/2bh
(5)^2 - (4)^2 = 1/2(4)(3)
25 - 16 = 1/2(12)
9 not equal to 6
C. Area of ABCD + Area of DGA = Area of GHIA
s^2 + 1/2bh = s^2
(5)^2 + 1/2(4)(3) = (4)^2
25 + 1/2(12) = 16
25 + 6 = 16
31 not equal to 16
D. Area of DEFG + Area of GHIA = Area of ABCD
s^2 + s^2 = s^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25
The answer is D.
Answer:
x intercept: 15/8 and yintercerpt is 3/2
Step-by-step explanation:
to find y intercept you have to put equation in standard form.
10y=-8x+15
y=--4/5x+3/2
x intercept is when y=0, so we plug in 0
0=-4/5x+3/2
4/5x=3/2
x=15/8