Answer:
The side length of the table is x + 5
Step-by-step explanation:
To solve this, you need only take the square root of the area:
So the side length of the table is x + 5
Angle MNO is 90, it’s just the top right corner
If f(z)=m^z - n^z <span>and f(1)=2, f(2)=8
then
m - n = 2 so m = n + 2
m^2 - n^2 = 8
substitute </span>m = n + 2 into m^2 - n^2 = 8
so
m^2 - n^2 = 8
(n + 2)^2 - n^2 = 8
n^2 + 4n + 4 - n^2 = 8
4n = 4
n = 1
m = n + 2 = 1 + 2 = 3
so f(z) = 3^z - 1^z
if z = 1 the f(1) = 3^1 - 1^1 = 2
if z = 2 the f(2) = 3^2 - 1^2 = 9 - 1 = 8
if z = 3 the f(3) = 3^3 - 1^2 = 27 - 1 = 26
answer
f(3) = 26
Answer:
-2,9
Step-by-step explanation:
Dont want to explain too long
Step-by-step explanation:
You can find the area of a right triangle the same as you would any other triangle by using the following formula:
A = (1/2)bh, where A is the area of the triangle, b is the length of the base and h is the height of the triangle; However, with a right triangle, it's much more convenient in finding its area if we utilize the lengths of the two legs (the two sides that are shorter than the longest side, the hypotenuse and that are perpendicular to each other and thus form the right angle of the right triangle), that is, since the two legs of a right triangle are perpendicular to each other, when we treat one leg as the base, then, consequently, we can automatically treat the length of the other leg as the height, and if we initially know the lengths of both legs, then we can then just plug this information directly into the area formula for a triangle to find the area A of the right triangle.
For example: Find the area of a right triangle whose legs have lengths of 3 in. and 4 in.
Make the 4 in. leg the base. Since the two legs of a right triangle are perpendicular to each other, then the length of the other leg is automatically the height of the triangle; therefore, plugging this information into the formula for the area of a triangle, we have:
A = (1/2)bh
= (1/2)(4 in.)(3 in.)
= (1/2)(12 in.²)
A = 6 in.² (note: in.² means square inches)
