The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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Answer:
area=6
perimeter=36
Step-by-step explanation:
Answer:
A. -4
Step-by-step explanation:
For solving for x intercepts analytically. You can set the the y in the equation to 0. So, 2x-3(0)=12, and solving for x will get you -4.
You can also solve graphically by plugging in the equation and looking at where it intercepts the x axis.
Answer:
38.7°
Step-by-step explanation:
We can solve the triangle using the trigonometrical ratios which may be expressed in the form SOA CAH TOA Where
SOA stands for
Sin Ф = opposite side/hypotenuses side
CAH stands for
Cosine Ф = adjacent side/hypotenuses side
TOA stands for
Tangent Ф = opposite side/adjacent side
The hypotenuse is the side facing the right angle while the opposite is the side facing the given angle.
Considering the triangle with respect to angle x, 8cm is the opposite side, 10cm is the adjacent side
hence
Tan x = 8/10
Tan x = 0.8
x = tan -1 0.8
= 38.7°
Answer:
x = 52
Step-by-step explanation:
This is a picture of a right angle, which is 90 degrees. We should first make an equation to represent this problem.
(x - 12) + 50 = 90
1. Subtract 50 from both sides of the equation.
(x - 12) = 40
2. Add 12 from both sides of the equation
x = 52
The value of x is 52.
