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Grace [21]
3 years ago
11

Is the point 3/5 and 4/5 corresponds to an angle in the unit circle what is the csc

Mathematics
1 answer:
Mariulka [41]3 years ago
8 0

<u>Answer:</u>

csc = 5/4

<u>Step-by-step explanation:</u>

Assuming that the two given points 3/5 and 4/5 are the two sides of a triangle and a right angle 0  with legs x and y and z as its hypotenuse.

Then using Pythagoras Theorem:

z^2=x^2+y^2

z^2=(\frac{3}{5} )^2+(\frac{4}{5} )^2

z^2 = \frac{9}{25} + \frac{16}{25}

z^2=\frac{25}{25}

z^2=1

Taking square root on both the sides:

z=1

Now, sin\alpha=\frac{4}{5}

and we know that csc = \frac{1}{sin\alpha }

Therefore, csc = 5/4.

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The number of pens sold at the school bookstore on Monday, p, increased 45% on Tuesday. The expression p+0.45p represents the nu
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p + (45/100) p                                   or       1.45 p

Step-by-step explanation:

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2 years ago
Look at the picture
IRISSAK [1]

Answer:

x = 10

Step-by-step explanation:

In the last step, you can see that the fraction \frac{1}{5} has been multiplied by its reciprocal \frac{5}{1}, making it cancel out. The reciprocal has been multiplied to both sides, so all you need to do is multiply \frac{6}{3} · \frac{5}{1} :

6 · 5 = 30

3 · 1 = 3

<u><em>So now you should have the fraction:</em></u>

x = \frac{30}{3}

<u><em>But, you can still simplify the fraction by dividing 30 by 3:</em></u>

x = 10

7 0
3 years ago
Evaluate the expression for x=5, y=-3 and z=7<br> 7 - 3y – 3z
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Answer:

Step-by-step explanation:

7 - 3(-3) - 3(7) = 7 + 9 - 21 = 16 - 21 = -5

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Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
3 years ago
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