![{34}^{2} + {b}^{2} = {42}^{2} \\ 1156 + {b}^{2} = 1764 \\ 1156 - 1156 + {b}^{2} = 1764 - 1156 \\ {b}^{2} = 608 \\ \sqrt{ {b}^{2} } = \sqrt{608} \\ b = 24.7](https://tex.z-dn.net/?f=%20%7B34%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20%7B42%7D%5E%7B2%7D%20%20%5C%5C%201156%20%2B%20%7Bb%7D%5E%7B2%7D%20%20%3D%201764%20%5C%5C%201156%20-%201156%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%201764%20-%201156%20%5C%5C%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20608%20%5C%5C%20%20%5Csqrt%7B%20%7Bb%7D%5E%7B2%7D%20%7D%20%20%3D%20%20%5Csqrt%7B608%7D%20%20%5C%5C%20b%20%3D%2024.7)
I used the Pythagorean formula, which is a² + b² = c².
a² and b² are the two side legs while c² is the hypotenuse.
One of the side is 42 - 8 = 34 cm.
Answer:
distance=90km
time=1.5h
speed of the bus=distance /time
speed of the bus=90km/1.5h
speed of the bus=60km/h
therefore if 1.5h=60km/h
then 7.2h=(7.2x60)/1.5 =288km
If the problem is referring to the equivalent logarithmic equation log (20 *27).
We can easily find and solve its equivalent expression using one of the many identities available in logarithmic.
We can have the expression:
log (20*27) = log 20 + log 27
Answer:
In this question if g and h is constant then;
111 = 14g
14g = 111
g = 111/14
g = 27.7 ~ 28
And also,
13.65 = h + 4.88
h + 4.88 = 13.65
h = 13.65 - 4.88
h = 8.77
Hope it helps!
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b
16=2⁴
32=2⁵
rememeber that
![(x^m)^n=x^{mn}](https://tex.z-dn.net/?f=%28x%5Em%29%5En%3Dx%5E%7Bmn%7D)
![16^{3x+2}=32^{-2x-7}](https://tex.z-dn.net/?f=16%5E%7B3x%2B2%7D%3D32%5E%7B-2x-7%7D)
![(2^4)^{3x+2}=(2^5)^{-2x-7}](https://tex.z-dn.net/?f=%282%5E4%29%5E%7B3x%2B2%7D%3D%282%5E5%29%5E%7B-2x-7%7D)
![2^{4(3x+2)}=2^{5(-2x-7)}](https://tex.z-dn.net/?f=2%5E%7B4%283x%2B2%29%7D%3D2%5E%7B5%28-2x-7%29%7D)
2=2 so we conclude that 4(3x+2)=5(-2x-7)
4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22