Answer:
1. Use the Adjacent and opposite side (Ignore the Hypotenuse)
Or use HERO'S FORMULA based on the information given
2. Area = 216cm^2
Step-by-step explanation:
There are three to four ways we can go about finding the area of a triangle. And a these would be dependent on the information given about the triangle.
From the question, you said the three side lengths are given. In such case, we employ the HERO FORMULA.
HERO FORMULA:
Area = √ s(s-a)(s-b)(s-c)
where s = 1/2(a + b + c)
a, b, c are the three sides
But since the question insisted that we use 1/2* base * height. Let's use our know of right angle to dissolve that.
A right angle triangle has three sides. The longest is always the Hypotenuse.
Let's take it this way.
Hypotenuse = 30cm
Opposite= 18cm
Adjacent = 24cm
Area = 1/2 * base * height
Area = 1/2 * 18 * 24
Area = 1/2 * 432
Area = 216cm^2
We ignored the longest side, (the Hypotenuse)
Volume of a cube = s³
Therefore, s³ = 512, meaning that s = 8.
The question asks what the resulting volume would be if all sides of the cube were divided by 4.
The current side measures 8, and we know that 8/4 = 2.
Thus, the resulting volume = s³ = 2³ = 8 cubic mm.
90=l+l+w+w
90=28+28+w+w
90=56+2w
34=2w
17=w
The width of the playground is 17.
Step-by-step explanation:
Answer above
if you have any doubts ask in comments
Given a solution

, we can attempt to find a solution of the form

. We have derivatives



Substituting into the ODE, we get


Setting

, we end up with the linear ODE

Multiplying both sides by

, we have

and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to

, we get


Now solve for

:


So you have

and given that

, the second term in

is already taken into account in the solution set, which means that

, i.e. any constant solution is in the solution set.