V1 - velocity first train
v2 - velocity second train
v2 > v1
v2 - v1 = 17 mph
We know, that:
![s_1+s_2=210 \ [miles] \\ \\ t_1=t_2=2h](https://tex.z-dn.net/?f=s_1%2Bs_2%3D210%20%5C%20%5Bmiles%5D%20%5C%5C%20%5C%5C%20t_1%3Dt_2%3D2h)
So:
NOw we've got simple system of equations:
![+\begin{cases} v_2-v_1=17 \\ v_2+v_1=105\end{cases} \\ \\ 2v_2=122 \qquad /:2 \\ \\ v_2=61 \qquad [mph] \\ \\ v_2-v_1=17 \\ \\ 61-v_1=17 \\ \\ v_1=44](https://tex.z-dn.net/?f=%2B%5Cbegin%7Bcases%7D%20v_2-v_1%3D17%20%5C%5C%20v_2%2Bv_1%3D105%5Cend%7Bcases%7D%20%5C%5C%20%5C%5C%202v_2%3D122%20%5Cqquad%20%2F%3A2%20%5C%5C%20%5C%5C%20v_2%3D61%20%5Cqquad%20%5Bmph%5D%20%5C%5C%20%5C%5C%20v_2-v_1%3D17%20%5C%5C%20%5C%5C%2061-v_1%3D17%20%5C%5C%20%5C%5C%20v_1%3D44)
Velocities of these trains are 61mph and 44mph
Answer:
The number of cases prior to the increase is 50.
Step-by-step explanation:
It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.
We need to find the number of cases prior to the increase.
Let x be the number of cases prior to the increase.
x + 13.6% of x = 57
Divide both the sides by 1.136.
Therefore the number of cases prior to the increase is 50.
Answer:
Step-by-step explanation:
Part A
<u>Givens</u>
b1 = 265
b2 = 180
h = 90
<u>Formula</u>
Area = (b1 + b2)*h / 2
<u>Solution</u>
Area = (265 + 180) * 90 / 2
Area = 445 * 90 / 2
Area = 40050 / 2
Area = 20025 square meters
Part B
1 hectare = 10000 square meters
x hectare = 20025 square meters Cross multiply
x * 10000 = 20025 * 1 Divide by 10000
x = 20025 / 10000
x = 2.0025
Rounded to the nearest hectare that would be 2
Part C
The best shape is a square.
That would mean that the area is given by
s^2 = 20025
sqrt(s^2) = sqrt(20025)
s = 141.51
Each side of the rectangle (square) = 141.51
The equation of line is y = mx+b ; m is slope , b is y- intercept
Please give brainlest
<h3>Answer:</h3>
A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°
<h3>Explanation:</h3>
The sum of angles in ∆ABC is 180°, so ...
... (2x -2) + (2x +2) + (5x) = 180
... 9x = 180
... x = 20
and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.
___
The sum of angles of ∆A'B'C' is 180°, so ...
... (58 -x) +(3x -18) +(120 -x) = 180
... x +160 = 180
... x = 20
and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.
_____
The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...
... A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°
