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Mazyrski [523]
3 years ago
6

In a topographical map of a city, the vertices of the city limits are A (10, 9), B (18, 9), C (18, 2), D (14, 4.5), and E (10, 4

.5). The coordinates are measured in miles. What is the area of the city?
Mathematics
1 answer:
harkovskaia [24]3 years ago
5 0
Srry don't have answer NEEEEEDDD HELP!!
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Someone help me please what is 4.3 x 20
Mars2501 [29]

Answer:

86

Step-by-step explanation:

7 0
3 years ago
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Which of the following statements is true about the division expression 686.54 ÷ a? A. If a is a number greater than 686.54, the
borishaifa [10]

Answer:

D-- if a is a number between 0 and 1, the quotient will be greater

Step-by-step explanation:

5 0
4 years ago
The results of a poll show that the percent of people in a town who want an observatory built on a nearby mountain is in the int
V125BC [204]
In finding this value you average lower and upper bound
  (0.6+0.82)/2 = 0.71   =0.71  estimated    margin of error = distance from  estimate point lower/ upper bound   This interval will be twice margin error
   (0.82-0.6)/2 = 0.11    How far is 0.82 from 0.71 ??   =0.11=11%

7 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Pythagorean Theorem
pshichka [43]

Answer:

1) 6^2 + 8^2 = x^2

36 + 64 = x^2

100 = x^2

x = 10 feet

2) 15^2 + 8^2 = x^2

225 + 64 = x^2

x ^ 2 = 289

guy rope = 17 feet


Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
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