Answer:
x = 0, x = - 3
Step-by-step explanation:
Given
5x² - 7x = 4x² - 10 ← subtract 4x² - 10x from both sides
x² + 3x = 0 ← in standard form
Factor out x from each term
x(x + 3) = 0
Equate each factor to zero and solve for x
x = 0
x + 3 = 0 ⇒ x = - 3
Let's have the first number, the larger number, be <em>x</em>. We'll have the second, smaller number be <em>y</em>.
We know that x = y + 6, since x is 6 greater than y.
We also know that 330 = x + y.
Because x = y + 6, 330 = y + 6 + y, which simplifies to 330 = 2y + 6.
Now all we need to do is simplify the equation. First, we subtract 6 from both sides:
330 - 6 = 324
2y + 6 - 6 = 2y.
So we have 324 = 2y. Then we divide both sides by 2 to get:
162 = y
Plug in y = 162 into the equation x = y + 6 to get:
x = 162 + 6
x = 168
Let's check to make sure our answer is right. 168 is 6 more than 162. 162 + 168 equals 330. So our two numbers are 168 and 162.
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:
r^2=(A/pi)
Step-by-step explanation:
if A=pi*r^2 then maybe dividing by pi to get r^2 alone says r^2=(A/pi) or to get rid of the exponent you may square root each side so r=(sqrt(A/pi))
