For this case we have that the distance between two points is given by the following formula:
![d = \sqrt {(x_ {2} -x_ {1})^ 2+(y_ {2} -y_ {1}) ^ 2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%20%7B%28x_%20%7B2%7D%20-x_%20%7B1%7D%29%5E%202%2B%28y_%20%7B2%7D%20-y_%20%7B1%7D%29%20%5E%202%7D)
We have the following points:
![(x_ {1}, y_ {1}): (-1,4)\\(x_ {2}, y_ {2}): (2,0)](https://tex.z-dn.net/?f=%28x_%20%7B1%7D%2C%20y_%20%7B1%7D%29%3A%20%28-1%2C4%29%5C%5C%28x_%20%7B2%7D%2C%20y_%20%7B2%7D%29%3A%20%282%2C0%29)
Substituting we have:
![d = \sqrt {(2 - (- 1)) ^ 2+ (0-4) ^ 2}\\d = \sqrt {(2 + 1) ^ 2 + (- 4) ^ 2}\\d = \sqrt {(3) ^ 2 + (- 4) ^ 2}\\d = \sqrt {9 + 16}\\d = \sqrt {25}\\d = 5](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%20%7B%282%20-%20%28-%201%29%29%20%5E%202%2B%20%280-4%29%20%5E%202%7D%5C%5Cd%20%3D%20%5Csqrt%20%7B%282%20%2B%201%29%20%5E%202%20%2B%20%28-%204%29%20%5E%202%7D%5C%5Cd%20%3D%20%5Csqrt%20%7B%283%29%20%5E%202%20%2B%20%28-%204%29%20%5E%202%7D%5C%5Cd%20%3D%20%5Csqrt%20%7B9%20%2B%2016%7D%5C%5Cd%20%3D%20%5Csqrt%20%7B25%7D%5C%5Cd%20%3D%205)
Thus, the distance between the points is 5 units.
Answer:
![d = 5](https://tex.z-dn.net/?f=d%20%3D%205)
x=12 and y=18 satisfies both the equalities.
Step-by-step explanation:
Given equations are;
x/y = 2/3 Eqn 1
y/24 = 3/4 Eqn 2
Multplying by 24 on both sides of Eqn 2
![24*\frac{y}{24}=\frac{3}{4}*24\\\frac{24y}{24}=\frac{72}{4}\\y=18](https://tex.z-dn.net/?f=24%2A%5Cfrac%7By%7D%7B24%7D%3D%5Cfrac%7B3%7D%7B4%7D%2A24%5C%5C%5Cfrac%7B24y%7D%7B24%7D%3D%5Cfrac%7B72%7D%7B4%7D%5C%5Cy%3D18)
Putting y=18 in Eqn 1;
![\frac{x}{18}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B18%7D%3D%5Cfrac%7B2%7D%7B3%7D)
Multiplying both sides by 18
![18*\frac{x}{18}=\frac{2}{3}*18\\\frac{18x}{18}=\frac{36}{3}\\x=12](https://tex.z-dn.net/?f=18%2A%5Cfrac%7Bx%7D%7B18%7D%3D%5Cfrac%7B2%7D%7B3%7D%2A18%5C%5C%5Cfrac%7B18x%7D%7B18%7D%3D%5Cfrac%7B36%7D%7B3%7D%5C%5Cx%3D12)
Putting both values in Eqn 1
![\frac{12}{18}=\frac{2}{3}\\](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B18%7D%3D%5Cfrac%7B2%7D%7B3%7D%5C%5C)
As 12 and 18 are multiples of 6;
![\frac{2}{3}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%3D%5Cfrac%7B2%7D%7B3%7D)
x=12 and y=18 satisfies both the equalities.
Keywords: fraction, division
Learn more about fractions at:
#LearnwithBrainly
Answer:
The correct answer is 24
Step-by-step explanation:
to solve this you will need to use the pathagreom theorum
a^{2}+b^{2}=c^{2}
A= one side lenth
B= the secons side lenth
C= hypotnuse
It is helpfull to draw out the situation
you know that the latter is 25 ft, that is your hypotnuse
you also know that the 7 ft away from the base of the building is one of the side lenths, lets call it side a
so plug the numbers into the equation
7^2 + b^2 = 25 ^2
you leave b^2 alone because that is the side you are trying to find
now square 7 and 25 but leave b^2 alone
49 + b^2 = 625
now subtract 49 from both sides
b^2 = 576
now to get rid of the square of b you have to do the opposite and square root both sides removing the square of the B and giving you the answer of..........
B= 24
Hope this helped!! I tryed to explain it as simpil as possiable
Answer:
![tan(30) = \frac{5}{b}](https://tex.z-dn.net/?f=tan%2830%29%20%3D%20%20%5Cfrac%7B5%7D%7Bb%7D%20)
Step-by-step explanation:
Trigonometric Ratios.
To solve for b, we check the parameters that are given in the triangle.
If we're considering 30°, we can see that the opposite is given as 5cm and the adjacent is b.
Applying:
![tan \alpha = \frac{oppsite}{adjacent} \\ \\ tan \: (30) = \frac{5}{b}](https://tex.z-dn.net/?f=tan%20%5Calpha%20%20%3D%20%20%5Cfrac%7Boppsite%7D%7Badjacent%7D%20%20%5C%5C%20%20%5C%5C%20tan%20%5C%3A%20%2830%29%20%3D%20%20%5Cfrac%7B5%7D%7Bb%7D%20)