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tankabanditka [31]
3 years ago
7

Enter an equation in standard form to model the linear situation.

Mathematics
1 answer:
sineoko [7]3 years ago
3 0

Answer:

9x - y = -12

Step-by-step explanation:

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Help please!!!!!!!!!!!!!
Dafna11 [192]

Answer: A and C

Step-by-step explanation: Adding A and C, we get (-3-4i)+(13+i) we get 10-3i, the desired sum. Hope this helps! (Could I please have brainliest?)

7 0
4 years ago
Can y’all please help me ……
Goryan [66]

Answer:

Step-by-step explanation:

<u><em>( 8 ).</em></u> m∠1 = m∠ 4 = m∠ 5 = <em>m∠6 = 52°</em>

<em><u>( 9 ).</u></em> m∠ 2 = m∠ 3 = <em>m∠ 7</em> = m∠ 8 = 180° - 52° <em>= 128°</em>

5 0
2 years ago
Pleeeaaasseeee help (questions in photo) will mark brainliest
KIM [24]

Answer:

a. 0.8727rad

b. 5.585rad

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5 0
2 years ago
What is the period of the graph of the equation y = a sin bx?
steposvetlana [31]

T=2π/|b|. The period of an equation of the form y = a sin bx is T=2π/|b|.

In mathematics the curve that graphically represents the sine function and also that function itself is called sinusoid or sinusoid. It is a curve that describes a repetitive and smooth oscillation. It can be represented as y(x) = a sin (ωx+φ) where a is the amplitude, ω is the angular velocity with ω=2πf, (ωx+φ) is the oscillation phase, and φ the initial phase.

The period T of the sin function is T=1/f, from the equation ω=2πf we can clear f and substitute in T=1/f.

f=ω/2π

Substituting in T=1/f:

T=1/ω/2π -------> T = 2π/ω

For the example y = a sin bx, we have that a is the amplitude, b is ω and the initial phase φ = 0. So, we have that the period T of the function a sin bx is:

T=2π/|b|

7 0
3 years ago
I need help again.. can someone help me solve this...? I didn't get a screenshot of how to solve the perimeter but I know it had
fenix001 [56]

Answer:

21

Step-by-step explanation:

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3 years ago
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