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Natalija [7]
2 years ago
9

A 29 foot ladder leans against a building so that the angle between the ground and the ladder is 77º. How

Mathematics
1 answer:
Dimas [21]2 years ago
7 0

Answer:

5 √ 10 + 2 √ 5 ft high

Step-by-step explanation:Explanation:

Let  h be the height to which latter of length of 20 ft reach then using Sine formula in right triangle with hypotenuse 20 & vertical leg h & opposite acute angle 0 = 72∘ sin θ =h 20 sin 72∘= h 20

h

=

20

sin

(

90

∘

−

18

∘

)

=

20

cos

18

∘

=

20

(

√

10

+

2

√

5

4

)

=

5

√

10

+

2

√

5

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Julio has started his very own floral business. He sells centerpieces for $19.50 each. His fixed costs are $500 per month, and e
Phantasy [73]
The total cost, C, is the sum of the cost to produce the centerpieces and the fixed cost. The equation for C is C = 8n + 500.

The total revenue, R, of Julio's floral business is the product of the cost per centerpiece and the number of centerpieces sold. The equation for R is R = 19.50n. 
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3 years ago
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I really need help I’m very confused
Vaselesa [24]

Answer:

10: -11n^5

12: 6k^2 -6k+7

Step-by-step explanation:

Ok so what your gonna do is add or subtract the ones with the same exponent.

so -15+4= -11 since they are both n^5 you can add them so your answer is:

-11n^2

now you have 8k^2-k-5k+7-2k. so you are gonna rearrange them so the same exponents are together.(keep the sign in front in front, if no sign it is positive)

8k^2-2k^2-5k-k+7

add like exponents

6k^2-6k+7

6 0
3 years ago
Multi answer inequality! Please help and comment only if you know!
bearhunter [10]

Answer:

-8, 0, and -12

Step-by-step explanation:

Using -7 would get you 8. Using -5 would get you 4

6 0
2 years ago
A/dc=b ,solve for a ---please help?!!
nydimaria [60]
Multiply dc over one on both sides on the left it’ll be cancelled out with a alone and on the right it would be b*dc - from a freshman
7 0
2 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
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