Answer:
Step-by-step explanation:
So we have the trinomial:
First, factor out a 2:
Now, factor within the parentheses.
To do so, we want to find two numbers. When multiplied, these two numbers must equal (a)(c) and when added, they must equal b
(a)(c) is 2(-5) or -10 and b is -3. So, we want two numbers that when multiplied together gives -10 and when added gives -3.
-5 and 2 works. Therefore:
Replace -3x with 2x and -5x:
Factor out a 2x for the first two terms. And factor out a negative 5 for the remaining two:
Grouping:
And we're done!
54cm is equivalent to 0.59 yards
To write the equation, you can start with the answer of $680. She receives $400 + $20*plans sold:
400 + 20p = 680
20p = 280
p = 14
{$400 + $20 * 14 = 680}
Sam sold 14 plans to make a total of $680.
Answer:
r=294.9m
Step-by-step explanation:
The forces on the particle are
Now , we sum all these forces to get the net force
we can use the fact F=m*a and integrate the acceleration
![a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]](https://tex.z-dn.net/?f=a%28t%29%3D%5Cfrac%7B1%7D%7Bm%7DF%28t%29%5C%5C%5C%5Cv%28t%29%3D%5Cint%20a%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5Cint%7BF_%7BT%7D%7Ddt%5C%5C%5C%5Cv%28t%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%2857t-t%5E%7B2%7D%2B%5Cfrac%7B5%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bi%7D%2B%2812t-2t%5E%7B2%7D%29%5Chat%7Bj%7D%2B%28-t%5E%7B2%7D-t%29%5Chat%7Bk%7D%5D%5C%5C%5C%5Cr%28t%29%3D%5Cint%20v%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5B%28%5Cfrac%7B57%7D%7B2%7Dt%5E%7B2%7D-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D%7D%2B%5Cfrac%7B5%7D%7B4%7Dt%5E%7B4%7D%29%5Chat%7Bi%7D%2B%286t%5E%7B2%7D-%5Cfrac%7B2%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bj%7D%2B%28-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D-%5Cfrac%7B1%7D%7B2%7Dt%5E%7B2%7D%29%5D)
and we evaluate in r(2) an we take the norm to obtain the distance
![r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m](https://tex.z-dn.net/?f=r%282%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%5Cfrac%7B394%7D%7B3%7D%5Chat%7Bi%7D%2B%5Cfrac%7B56%7D%7B3%7D%5Chat%7Bj%7D-%5Cfrac%7B14%7D%7B3%7D%5Chat%7Bk%7D%5D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B1%7D%7Bm%7D%5Csqrt%7B%5B%28%5Cfrac%7B394%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B56%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B14%7D%7B3%7D%29%5E%7B2%7D%5D%7D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B132.73%7D%7B0.45%7D%3D294.9m)
I hope this is useful for you
regards
The answer is B because the difference from 20 hrs. of training to 10 hrs. of training is 100 dollars.
