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Igoryamba
2 years ago
6

Please help!! i’ll mark you as brainliest!!

Mathematics
1 answer:
PtichkaEL [24]2 years ago
5 0

Answer:

For Triangle 1

a. Hypothenus = 4

b. Adjacent = 2√3

For Triangle 2:

a. Hypothenus = 2√2

b. Adjacent = 2

Step-by-step explanation:

For Triangle 1

NOTE: We shall be using angle 30°.

a. Determination of the Hypothenus.

Opposite = 2

Angle (θ) = 30°

Hypothenus =?

Sine θ = Opposite /Hypothenus

Sine 30 = 2 / Hypo

0.5 = 2 / Hypo

Cross multiply

0.5 × Hypo = 2

Divide both side by 0.5

Hypo = 2 / 0.5

Hypothenus = 4

b. Determination of the Adjacent

NOTE: We shall be using angle 30°.

Opposite = 2

Angle (θ) = 30°

Adjacent =?

Tan θ = Opposite / Adjacent

Tan 30 = 2 / Adj

1 / √3 = 2 / Adj

Cross multiply

1 × Adj = 2 × √3

Adjacent = 2√3

For Triangle 2:

a. Determination of the Hypothenus.

Opposite = 2

Angle (θ) = 45°

Hypothenus =?

Sine θ = Opposite /Hypothenus

Sine 45 = 2 / Hypo

1 / √2 = 2 / Hypo

Cross multiply

1 × Hypo = 2 × √2

Hypothenus = 2√2

b. Determination of the Adjacent

Opposite = 2

Angle (θ) = 45°

Adjacent =?

Tan θ = Opposite / Adjacent

Tan 45 = 2 / Adj

1 = 2 / Adj

Cross multiply

1 × Adj = 2

Adjacent = 2

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Rom4ik [11]

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Step-by-step explanation:

in math is means either less than, greater than , or equal to

5 0
3 years ago
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

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

so the characteristic solution is

y_c=C_1e^{-t}+C_2te^{-t}

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
2 years ago
Find two different values that complete each expression so that the trinomial can be factored into the product of two binomials.
zalisa [80]
4s² + <u>14</u>s + 10
4s² + 4s + 10s + 10
2s(2s) + 2s(2) + 5(2s) + 5(2)
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(2s + 5)(2s + 2)

4s² + <u>22</u>s + 10
4s² + 20s + 2s + 10
2s(2s) + 2s(10) + 1(2s) + 1(10)
2s(2s + 10) + 1(2s + 10)
(2s + 1)(2s + 10)

It could be equal to 14 or 22.
3 0
3 years ago
Solve this please and explain the steps i dont get it
sashaice [31]

x > -60


1.) Multiply -10 and 6 so you get rid of the fraction

2.) Simply whats left

8 0
2 years ago
PLEASE please help i’ll give brainliest
elena55 [62]

Answer:

Step-by-step explanation:

The correct answer is dotplot

8 0
3 years ago
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