Answer:
x = 4, y = 2
Step-by-step explanation:
Given quadrilateral is a parallelogram. Diagonals of a parallelogram bisects each other.

Just has to be greater than or equal to 0... w and t can never be negative since taking the square root of a negative number yields an imaginary number which would mean imaginary current. Also the question states that w can't be zero but if T is zero than the current will be zero.
The formula to find the slope of a line is m =

where the x's and y's are your given coordinates and m is your slope. So, plug in your coordinates and solve.
m = <span>
![\frac{y_2 - y_1}{x_2 - x_1} Plug in your coordinates m = [tex] \frac{-2 - 7}{8 - -1} Cancel out the double negative m = [tex] \frac{-2 - 7}{8 + 1} Simplify m = [tex] \frac{-9}{9} Divide m = -1 Now, plug that slope and one set of your given coordinates into point-slope form, [tex]y - y_1 = m(x - x_1)](https://tex.z-dn.net/?f=%20%5Cfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D%20%20%20Plug%20in%20your%20coordinates%20%3C%2Fspan%3Em%20%3D%20%3Cspan%3E%5Btex%5D%20%5Cfrac%7B-2%20-%207%7D%7B8%20-%20-1%7D%20%20%20Cancel%20out%20the%20double%20negative%20%3C%2Fspan%3Em%20%3D%20%3Cspan%3E%5Btex%5D%20%5Cfrac%7B-2%20-%207%7D%7B8%20%2B%201%7D%20%20%20Simplify%20%3C%2Fspan%3Em%20%3D%20%3Cspan%3E%5Btex%5D%20%5Cfrac%7B-9%7D%7B9%7D%20%20%20Divide%20m%20%3D%20-1%20%20%3C%2Fspan%3E%20Now%2C%20plug%20that%20slope%20and%20one%20set%20of%20your%20given%20coordinates%20into%20point-slope%20form%2C%20%5Btex%5Dy%20-%20y_1%20%3D%20m%28x%20-%20x_1%29)
. I'll use (-1, 7).
<span>

Plug in your points and slope
</span>y - 7 = -1(x - -1) Cancel out the double negative
y - 7 = -1(x + 1) Use the Distributive Property
y - 7 = -x - 1 Add 7 to both sides
y = -x + 6
</span>
Answer:
14-3n
Step-by-step explanation:
Step-by-step explanation:
<u>Properties used</u>
- logₐ a = 1
- log aᵇ = b log a
- log ab = log a + log b
See the steps below
- - log (4*10⁻³) =
- - (log 4 + log 10⁻³) = (log 4 ≈ 0.6 rounded)
- - (0.6 - 3*log 10) =
- - (0.6 - 3*1) =
- - (0.6 - 3) =
- - (- 2.4) =
- 2.4