For this case we have a square whose sides are known and equal to 60 ft.
We want to find the diagonal of the square.
For this, we use the Pythagorean theorem.
We have then:
Answer:
from home to second base it is about:
![d = 60\sqrt{2} feet](https://tex.z-dn.net/?f=d%20%3D%2060%5Csqrt%7B2%7D%20feet)
Answer: 7+5(7)+8(4)-2
Step-by-step explanation:
that’s the answer
Answer:
y=40
Step-by-step explanation:
32 divided by 4 equals 8. And 5 times 8 equals 40.
Answer:
Step-by-step explanation:
<u>Given expression:</u>
<u>Find its value when x = 4 and y = 1, substitute:</u>
- 3(4 + 1) + 2*4² + (1 + 1)² =
- 3*5 + 2*16 + 2² =
- 15 + 32 + 4 =
- 51
Answer:
0.012
Step-by-step explanation:
Linear approximation says that,
![f(x) \approx f(x_o)+f'(x_o)(x-x_o)](https://tex.z-dn.net/?f=f%28x%29%20%5Capprox%20f%28x_o%29%2Bf%27%28x_o%29%28x-x_o%29)
For a cube the surface area is
.
So the side is 1.0 inch in, the surface area is
square inches.
In Linear approximation means you ignore the term
, if
is a small number, because then
will be a very smalle number and that does not contribute much to the error.
So the surface area is approximately,
![6x^2=6x_o^2+12x_o(x-x_o)](https://tex.z-dn.net/?f=6x%5E2%3D6x_o%5E2%2B12x_o%28x-x_o%29)
So here, ![x=1.001, x_o=0.001](https://tex.z-dn.net/?f=x%3D1.001%2C%20x_o%3D0.001)
The error in the area is approximately,
![12 \times 0.001=0.012](https://tex.z-dn.net/?f=12%20%5Ctimes%200.001%3D0.012)
So the error is 0.012.