Answer:
6 * 10^2
Step-by-step explanation:
Simplify : (4.2×10^7)÷(7×10^4)
(4.2 × 10^7) / (7 × 10^4)
From indices : 10^a ÷ 10^b = 10^(a - b)
(4.2 / 7) * 10^(7-4)
0.6 * 10^3
= 6 * 10^2
Hence, (4.2×10^7)÷(7×10^4) = 6 * 10^2
Answer:
4 2/3
Step-by-step explanation:
Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.
Answer:
D=100(opposites angles are equal)
A=C=80 (adj angles are supplementry)
Step-by-step explanation:
ANSWER
The correct answer is C
EXPLANATION
We want to find the quotient:

We multiply by the reciprocal of the second fraction:

We cancel out the common factors to obtain:

We multiply to get

This simplifies to :

The correct answer is C