Y ^(-2/3)^-6 = y^12/3 = y^4
x^(1/2)^ -6 = x^-3 = 1 x^3
so the answer is y^4 * 1 /x^3 = y^4 / x^3
A
Answer:
Step-by-step explanation:
Problem One
All quadrilaterals have angles that add up to 360 degrees.
Tangents touch the circle in such a way that the radius and the tangent form a right angle at the point of contact.
Solution
x + 115 + 90 + 90 = 360
x + 295 = 360
x + 295 - 295 = 360 - 295
x = 65
Problem Two
From the previous problem, you know that where the 6 and 8 meet is a right angle.
Therefore you can use a^2 + b^2 = c^2
a = 6
b =8
c = ?
6^2 + 8^2 = c^2
c^2 = 36 + 64
c^2 = 100
sqrt(c^2) = sqrt(100)
c = 10
x = 10
Problem 3
No guarantees on this one. I'm not sure how the diagram is set up. I take the 4 to be the length from the bottom of the line marked 10 to the intersect point of the tangent with the circle.
That means that the measurement left is 10 - 4 = 6
x and 6 are both tangents from the upper point of the line marked 10.
Therefore x = 6
Slope = (5+7)/(1-4) = 12/-3 = -4
slope = -4
y = mx + b
-7 = -4(4) + b
-7 = -16 + b
b = 9
equation
y = -4x + 9
Answer:
d=5
Step-by-step explanation:
1. use the distance formula: d=
2. plug in the points: d=
3. solve problem inside parentheses: d=
4. square -4 and -3: d=
5. add:d=
6. find square root: d=5
Answer:
a=11
Step-by-step explanation:
