Answer: 37 units
Step-by-step explanation:
This also works as the height of the triangle.
This also works as the base of the triangle.
Let's call pink ''a'', and blue ''b''. The side we're looking for ''c'' is the hypothenuse.
To find the values of a and b, use the area formula of a square and solve for a side. In this case, since we're going to need the squared values, this step can be omitted.

![s=\sqrt[]{A}](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7BA%7D)
Let's work with Blue.
![s=\sqrt[]{144units^2} \\s=12units](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7B144units%5E2%7D%20%5C%5Cs%3D12units)
Now Pink.
![s=\sqrt[]{1225units^2}\\s=35units](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7B1225units%5E2%7D%5C%5Cs%3D35units)
So we have a triangle with a base of 35 units and a height of 12 units.
Now let's use the pythagoream's theorem to solve.
![c^2=a^2+b^2\\c=\sqrt[]{a^2+b^2} \\c=\sqrt[]{(12units)^2+(35units)^2}\\c=\sqrt[]{144units^2+1225units^2}\\ c=\sqrt[]{1369units^2}\\ c=37units](https://tex.z-dn.net/?f=c%5E2%3Da%5E2%2Bb%5E2%5C%5Cc%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%20%5C%5Cc%3D%5Csqrt%5B%5D%7B%2812units%29%5E2%2B%2835units%29%5E2%7D%5C%5Cc%3D%5Csqrt%5B%5D%7B144units%5E2%2B1225units%5E2%7D%5C%5C%20c%3D%5Csqrt%5B%5D%7B1369units%5E2%7D%5C%5C%20c%3D37units)
Answer:
aₙ= 6 n
Step-by-step explanation:
A honey comb cell contain 6 cells
a₁ = 6
d=6
aₙ = 6 + (n-1)6
= 6 n
Answer:
Stewhapt
rectangles-by-step explanation:
We have been given a table of overnight low temperature for a four-day period. We are supposed to give a possible temperature for Wednesday.
We are told that the low temperature on Wednesday was between the second and third highest temperature.
Let us place our given temperatures from least to greatest order.
Sunday (-11), Thursday (-8), Friday (7), Sunday (18)
We can see that second and third highest temperatures are -8 and 7 respectively.
Any number between -8 and 7 will be a possible temperature for Wednesday. 3 can be a possible temperature for Wednesday as 3 is between -8 and 7.
y ≤ x − 5
y ≥ −x − 4
This is a check and see type of problem.
if x=5
y ≤ 5 − 5 y≤ 0
y ≥ −5 − 4
y ≥-9
-9≤ y ≤ 0
-2 works so (5,-2) is a solution
if x= -5
y ≤ -5 − 5 y≤ -10
y ≥ 5 − 4
y ≥1
no solution
Choice B (5,-2)