The answer is the last one
Your answer should be -8y + 6.6z + 10
Answer:
Step-by-step explanation:
Given:
Area = 3x^3 - 16x^2 + 31x - 20
Base:
x^3 - 5x
Area of trapezoid, S = 1/2 × (A + B) × h
Using long division,
(2 × (3x^3 - 16x^2 + 31x - 20))/x^3 - 5x
= (6x^3 - 32x^2 + 62x - 40))/x^3 - 5x = 6 - (32x^2 - 92x + 40)/x^3 - 5x = 2S/Bh - Ah/Bh
= 2S/Bh - A/B
= (2S/B × 1/h) - A/B
Since, x^3 - 5x = B
Comparing the above,
A = 32x^2 - 92x + 40
2S/B = 6
Therefore, h = 1
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
45 dogs and 52 cats because 5+7=12 so 45+7=52
