Answer:
x = 32
m∠7 = 94
m∠8 = 86
m∠3 = 94
Step-by-step explanation:
(2x + 30) + (3x - 10) = 180
5x + 20 = 180
5x = 160
x = 32
m∠7 = 2(32) + 30 = 94
m∠8 = 3(32) - 10 = 86
<u>or</u>
m∠8 = 180 - 94 = 86
<u>m∠3 ≈ m∠7</u> (corresponding angles)
Answer:
The roots are
![x=\frac{1}{6} [3+ i\sqrt{3}]](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B6%7D%20%5B3%2B%20i%5Csqrt%7B3%7D%5D)
![x=\frac{1}{6} [3- i\sqrt{3}]](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B6%7D%20%5B3-%20i%5Csqrt%7B3%7D%5D)
Step-by-step explanation:
we have
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
substitute in the formula
Remember that
so
![x=\frac{1}{6} [3\pm i\sqrt{3}]](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B6%7D%20%5B3%5Cpm%20i%5Csqrt%7B3%7D%5D)
The roots are
Answer:
Step-by-step explanation:
Explanation:
We are not given units for distance or speed. The only units we're given are for time (seconds). Let the car's speed be represented by "v".
When the truck is moving 5 times as fast as the car, its relative speed is ...
5v -v = 4v
For some time (t) at the higher speed, it will move a distance farther than the car moves. That distance is the product of the relative speed and the corresponding time period: 4vt. Since the speeds are in the same direction, this additional distance is added to the initial relative location of the truck, -60. The new position of the truck relative to the car is ...
-60 +4vt = -60 +4v(8 seconds)
-60 + 32v . . . . . new position of the truck in relation to the car
(This only makes sense if the units of v are the same distance units as -60 and time units of seconds.)
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At least, operations of multiplication and addition are needed. Depending on the units of speed and distance, units conversion may also be required.
