1)what equation represents a line that is perpendicular to the line whose equation is -2y=3x+7?
2)if the measures of the angles of a triangle are reprsented by 2x, 3x-15, and 7x+15, is the triangle an acute triangle?
1. y=3/2x+7
so the equation is y=-2/3x+7
sorry, not sure about 2nd question
1) To find the equation of a line that is perpendicular to another line, we need to know that perpendicular lines have slopes that are negative reciprocals of each other.
Given the equation of the line:
-2y = 3x + 7
First, we rewrite it in slope-intercept form (y = mx + b) by dividing both sides by -2:
y = (-3/2)x - 7/2
The slope of this line is -3/2.
To find the slope of the perpendicular line, we take the negative reciprocal of -3/2, which means we invert the fraction and change the sign:
Slope of the perpendicular line = 2/3
Now, we have the slope of the perpendicular line. To find the equation, we can use the point-slope form of a line: y - y₁ = m(x - x₁), where (x₁, y₁) is any point on the line.
Let's say we want to find the equation passing through the origin (0,0):
y - 0 = (2/3)(x - 0)
Simplifying, we get:
y = (2/3)x
Therefore, the equation of the line perpendicular to -2y = 3x + 7 is y = (2/3)x.
2) To determine if a triangle is acute or not, we need to check if all three angles are less than 90 degrees.
Given the measures of the angles:
2x, 3x - 15, and 7x + 15
We can write the inequality for each angle:
2x < 90 (Angle A)
3x - 15 < 90 (Angle B)
7x + 15 < 90 (Angle C)
Now, we solve each inequality:
2x < 90
Divide both sides by 2:
x < 45
3x - 15 < 90
Add 15 to both sides:
3x < 105
Divide both sides by 3:
x < 35
7x + 15 < 90
Subtract 15 from both sides:
7x < 75
Divide both sides by 7:
x < 10.71
Now, we compare the values of x obtained from each inequality. As long as x is less than 10.71, all three angles will be less than 90 degrees. Therefore, the triangle is acute.