In order to begin we must start off with the formula for the area of a triangle, which is a=1/2b(h) where a is area, b is base, and h is height.
In this scenario, we know that the area is 45cm^2 and the base is 2h+12 (since it is twice it’s height plus twelve). We can plug this into the area equation and then proceed to solve out accordingly.
a=1/2b(h)
45=1/2(2h+12)(h)
90=(2h+12)(h)
90=2h^2 + 12h
0= 2h^2 + 12h - 90
Simplify by dividing the two out.
h^2 + 6h - 45 = 0.
Now plug into the quadratic formula (with a=1, b=6, and c=-45) as shown in the image below.
After plugging the equation in and solving, we come to the idea that h is roughly equal to 4.34. We can now plug this back into the triangle area formula to solve out for b.
a=1/2b(h)
45=1/2(2h + 12)(h)
45=1/2(20.69)(4.34)
45=45.
In conclusion;
The height is ≈ 4.34
The base is ≈ 8.68
Hope this helps :)
Teresa is correct because he rode a 1/4 mile bike trail 8 times and if you simplify 8/4 miles it would be 2 miles
Error// Error// Error// Error// Error// Error//
Answer: 0.8% of 5,700 = <u><em>45.6 </em></u>
Step-by-step explanation:
0.8% × 5,700 =
(0.8 ÷ 100) × 5,700 =
(0.8 × 5,700) ÷ 100 =
4,560 ÷ 100 =
45.6;
Since
, we can rewrite the integral as

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

Both integrals are quite immediate: you only need to use the power rule

to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.