Answer:
-1
Step-by-step explanation:
-7=n-6
n=-7+6
n=-1
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Answer:
The correct answer is 16w.
Step-by-step explanation:
w = 1w
1⋅w⋅16 =
1⋅1w⋅16 =
16w
Therefore, the correct answer is 16w.
Hope this helps! :D
Answer:

Step-by-step explanation:
We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.
We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is
.
Upon substituting our given values, we will get:

Let us solve for r.


We know that area of an equilateral triangle is equal to
, where s represents side length of triangle.

The area within circle and outside the triangle would be difference of area of circle and triangle as:

We can make a common denominator as:


Therefore, our required expression would be
.
Answer:
- a. the y-intercept of f(x) is greater than the y-intercept of g(x)
Step-by-step explanation:
<u>Given function:</u>
<u>We can work out the function from the table values:</u>
<u>Options:</u>
a. the y-intercept of f(x) is greater than the y-intercept of g(x)
- True. x= 0 ⇒ f(0) = (-4)^2 - 4 = 12, g(0) = 2^2 - 4 = 0
b. The y-intercept of g(x) is greater than the y-intercept of f(x)
- False. See previous option.
c. The y-coordinate of the vertex of f(x) s greater than the y-coordinate of the vertex of g(x)
- False. They are same = -4
d. The x-coordinate of the vertex of g(x) is greater than the x-coordinate of the vertex of f(x)