30 60 90 right triangle
so
ratio of short leg : long leg : hypo = a : a√3 : 2a
if short leg = 10
long leg = 10√3 and hypo = 2 x 10 = 20
so a = 10√3, c + d = 20
also small triangle on right is 30 60 90 right triangle
hypo = 10, short leg d = 5 and long leg b = 5√3
if d = 5 then c = 20 - 5 = 15
answer
a = 10√3
b = 5√3
c = 15
d = 5
Answer: it’s a
Step-by-step explanation:
To solve the two equations simultaneously using the substitution method we need to rearrange one of the equation to make either
!['x'](https://tex.z-dn.net/?f=%27x%27)
or
!['y'](https://tex.z-dn.net/?f=%27y%27)
the subject.
We can try in turn rearranging both equations and see which unknown term would have been easier to solve first
Equation
![2x+8y=12](https://tex.z-dn.net/?f=2x%2B8y%3D12)
Making
!['x'](https://tex.z-dn.net/?f=%27x%27)
the subject
![2x=12-8y](https://tex.z-dn.net/?f=2x%3D12-8y)
, dividing each term by 2
![x=6-4y](https://tex.z-dn.net/?f=x%3D6-4y)
⇒ (Option 1)
Making
!['y'](https://tex.z-dn.net/?f=%27y%27)
the subject
![8y=12-2x](https://tex.z-dn.net/?f=8y%3D12-2x)
, multiply each term by 8 gives
![y= \frac{12}{8} - \frac{2}{8}x](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B12%7D%7B8%7D%20-%20%5Cfrac%7B2%7D%7B8%7Dx%20)
⇒ (Option 2)
Equation
![3x-8y=11](https://tex.z-dn.net/?f=3x-8y%3D11)
Making
!['x'](https://tex.z-dn.net/?f=%27x%27)
the subject
![3x=11+8y](https://tex.z-dn.net/?f=3x%3D11%2B8y)
, divide each term by 3
![x= \frac{11}{3}+ \frac{8}{3}y](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B11%7D%7B3%7D%2B%20%5Cfrac%7B8%7D%7B3%7Dy%20)
⇒ (Option 3)
Making
!['y'](https://tex.z-dn.net/?f=%27y%27)
the subject
![8y=3x-11](https://tex.z-dn.net/?f=8y%3D3x-11)
, divide each term by 8
![y= \frac{3}{8}x- \frac{11}{8}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B3%7D%7B8%7Dx-%20%5Cfrac%7B11%7D%7B8%7D%20)
⇒ (Option 4)
From all the possibilities of rearranged term, the most efficient option would have been the first option, from equation
![2x+8y=12](https://tex.z-dn.net/?f=2x%2B8y%3D12)
with
!['x'](https://tex.z-dn.net/?f=%27x%27)
as the subject,
Answer:
Yes
Step-by-step explanation:
If a triangle is a right triangle, the 3 side lengths will check out in the Pythagorean Theorem.
![a^2+b^2=c^2](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3Dc%5E2)
where a and b are the legs and c is the hypotenuse.
The legs are the 2 shorter lengths and the hypotenuse is the longest length. The 3 side lengths are: 16,63 and 65. Therefore, 16 and 63 are the legs and 65 is the hypotenuse.
a=16
b=63
c=65
![16^2+63^2=65^2](https://tex.z-dn.net/?f=16%5E2%2B63%5E2%3D65%5E2)
Evaluate each exponent.
16^2=16*16=256
![256+63^3=65^2](https://tex.z-dn.net/?f=256%2B63%5E3%3D65%5E2)
63^2=63*63=3969
![256+3969=65^2](https://tex.z-dn.net/?f=256%2B3969%3D65%5E2)
65^2=65*65=4225
![256+3969= 4225](https://tex.z-dn.net/?f=256%2B3969%3D%204225)
Add 256 and 3969
![4225=4225](https://tex.z-dn.net/?f=4225%3D4225)
The statement above is true; 4225 is equal to 4225. Therefore, this is a right triangle because the side lengths check out when plugged into the Pythagorean Theorem.