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Alona [7]
3 years ago
7

A right triangle has side lengths 9, 40, and 41 as shown below.

Mathematics
2 answers:
Mashutka [201]3 years ago
5 0

Answer:

cos B = 9/41

tan B = 40/9

sin B = 40/41

Step-by-step explanation:

There is no picture, so I am making a few assumptions.

I am assuming that angle C is the right angle and is opposite 41.

I am also assuming that angle B is formed by the sides that measure 9 and 41. The hypotenuse is 41.

For angle B, the adjacent leg is 9, and the opposite leg is 40.

cos B = adj/hyp = 9/41

tan B = opp/adj = 40/9

sin B = opp/hyp = 40/41

Anna11 [10]3 years ago
4 0
COS B= 9/41
tan B= 40/9
sin B= 40/41
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5 0
3 years ago
Pls help me on this question.
PIT_PIT [208]

Answer:

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3 0
3 years ago
Y''-2y'+y=2(e^x)-3(e^-x)
labwork [276]
First find the characteristic solution. The characteristic equation is

r^2-2r+1=(r-1)^2=0

which as one root at r=1 of multiplicity 2. This means the characteristic solution for this ODE is

y_c=C_1e^x+C_2xe^x

For the nonhomogeneous part, you can try a particular solution of the form

y_p=(a_2x^2+a_1x+a_0)e^x+be^{-x}

which has derivatives

{y_p}'=(a_2x^2+(2a_2+a_1)x+a_1+a_0)e^x-be^{-x}
{y_p}''=(a_2x^2+(4a_2+a_1)x+2a_2+2a_1+a_0)e^x+be^{-x}

Substituting into the ODE, the left hand side reduces significantly to

2a_2e^x+4be^{-x}=2e^x-3e^{-x}

and it follows that

\begin{cases}2a_2=2\\4b=-3\end{cases}\implies a_2=1,b=-\dfrac34

Therefore the particular solution is

y_p=x^2e^x-\dfrac34e^{-x}

and so the general solution is the sum of the characteristic and particular solutions,

y=y_c+y_p
y=C_1e^x+C_2xe^x+x^2e^x-\dfrac34e^{-x}
3 0
3 years ago
Find the missing angle in the triangle below
Makovka662 [10]
The answer would be 58

180 - 65 - 57 = 58 which is the missing angle
8 0
3 years ago
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Which of the following is the same as 2.3 × 10^3?
frutty [35]
The answer is 2300
Hope this help
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4 years ago
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