Answer:
disagree
Step-by-step explanation:
toa=opposit over adjacent= 12/9
tan(x)=12/9
tan^-1=12/9
x= 53.1
Answer:
10.3cm=0.103
Step-by-step explanation:
Hope this helped :)
-<em>El</em>
<h3>
<u>Answer:</u></h3>
<h3>
<u>Step-by-step explanation:</u></h3>
Here , two circles are given which are concentric. The radius of larger circle is 10cm and that of smaller circle is 4cm . And we need to find thelarea of shaded region.
From the figure it's clear that the area of shaded region will be the difference of areas of two circles.
Let the,
- Radius of smaller circle be r .
- Radius of smaller circle be r .
- Area of shaded region be
![\bf \implies Area_{Shaded}= Area_{bigger}-Area_{smaller} \\\\\bf\implies Area_{Shaded} = \pi R^2 - \pi r^2 \\\\\bf\implies Area_{shaded} = \pi ( R^2-r^2) \\\\\bf\implies Area_{shaded} = \pi [ (10cm)^2 - (4cm)^2] \\\\\bf\implies Area_{shaded} = \pi [ 100cm^2-16cm^2] \\\\\bf\implies Area_{shaded} = \pi \times 84cm^2 \\\\\bf\implies Area_{shaded} = \dfrac{22}{7}\times 84cm^2 \\\\\bf\implies \boxed{\red{\bf Area_{shaded} = 264 cm^2 }}](https://tex.z-dn.net/?f=%5Cbf%20%5Cimplies%20Area_%7BShaded%7D%3D%20Area_%7Bbigger%7D-Area_%7Bsmaller%7D%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7BShaded%7D%20%3D%20%5Cpi%20R%5E2%20-%20%5Cpi%20r%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%3D%20%5Cpi%20%28%20R%5E2-r%5E2%29%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%3D%20%5Cpi%20%5B%20%2810cm%29%5E2%20-%20%284cm%29%5E2%5D%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cpi%20%5B%20100cm%5E2-16cm%5E2%5D%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cpi%20%5Ctimes%2084cm%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cdfrac%7B22%7D%7B7%7D%5Ctimes%2084cm%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20%5Cboxed%7B%5Cred%7B%5Cbf%20Area_%7Bshaded%7D%20%3D%20264%20cm%5E2%20%7D%7D)
<h3>
<u>Hence </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> the</u><u> </u><u>shaded </u><u>region</u><u> is</u><u> </u><u>2</u><u>6</u><u>4</u><u> </u><u>cm²</u><u>.</u></h3>
The correct question is:
Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)
Answer:
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
Step-by-step explanation:
We need to verify that
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
We differentiate
x = c1e^(-t) + c2e^(3t)
twice in succession, and substitute the values of x, x', and x'' into the differential equation
x''- 2x' - 3x = 0
and see if it is satisfied.
Let us do that.
x = c1e^(-t) + c2e^(3t)
x' = -c1e^(-t) + 3c2e^(3t)
x'' = c1e^(-t) + 9c2e^(3t)
Now,
x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]
= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)
= 0
Therefore, the differential equation is satisfied, and hence, x is a solution.
Answer:
None
Step-by-step explanation:
Look at the map
