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IrinaK [193]
3 years ago
6

Solve this please help me

Mathematics
2 answers:
densk [106]3 years ago
7 0

Answer:

1/2 or 0.5

Hope it helps!

Verizon [17]3 years ago
5 0
Answer is
X=0.5,x=2^-1
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Ocean tides can be modeled by a sinusoidal function. Suppose that there is a low and high tide every 12 hours, and that high tid
USPshnik [31]

Answer:

(a)y=5sin(\frac{\pi}{6}(x+2))+5

(b) 7.5ft above the low tide.

Step-by-step explanation:

(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:

y=Asin(B(x-C))+D

Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.

The  amplitude is half the distance between the highest and the lowest tide:

A=10/2=5ft

The frequency is related to the period T by:

B=\frac{2\pi}{T}

The period is 12 hours, then

B=\frac{2\pi}{12}=\frac{\pi}{6}

The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when sin(B(x-C))=1, if sin(\frac{\pi}{2})=1 then B(x-C) must be equal to \frac{\pi}{2} when x=1:

B(x-C)=\frac{\pi}{2}\\\frac{\pi}{6}(1-C)=\frac{\pi}{2}\\\frac{1}{6}(1-C)=\frac{1}{2}\\(1-C)=\frac{6}{2}\\-C=3-1\\C=-2

The vertical shift is the sum of the lowest value, the height of the low tide (lt) and the amplitude:

D=5+lt

The function is:

y=5sin(\frac{\pi}{6} (x+2))+5+lt

Because the function must be the height above low tide height, subtract this heigh from the function:

y=5sin(\frac{\pi}{6} (x+2))+5+lt-lt

y=5sin(\frac{\pi}{6} (x+2))+5

(b) Use x=11 in the function

y=5sin(\frac{\pi}{6} (11+2))+5=2.5+5=7.5ft above the low tide.

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3 years ago
Jenny wants to buy an MP3 player that costs $30.98. How much change does she receive if she gives the cashier $40? PLEASE SHOW A
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40.00

-30.98

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9.02

You have to do some trades or borrowing. But think about it this way:

You need 2 cents to make it to $31. Then it’s $9 to get to 40.

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3 years ago
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Which word is used to name each of the bold numbers and variables?
eduard
The answer is b. Product
5 0
4 years ago
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Please HELPPPP me !!!!!
dedylja [7]

Answer:

y = 4x - 3\\Slope \ m_1 = 4

EQUATION OF LINE PERPENDICULAR

Lines \ are \  perpendicular => m_1 \cdot m_2 = -1\\

slope, m_2 = \frac{-1}{4}

equation \ of \ the\ line \ passing \ through \ (-2, 2) \ slope ,m_2 :\\(y - y_2) =m_2(x - x_2)\\ (y - 2) = \frac{-1}{4} (x -(-2))\\\\y - 2 = \frac{-1}{4} (x +2)\\\\y = \frac{-1}{4}x - \frac{1}{2}  + 2\\\\y = \frac{-1}{4}x + \frac{3}{2}

EQUATION OF LINE PARALLEL

Lines \ are \ parallel => m_1 \cdot m_3 = 1\\slope, m_3 = \frac{1}{4}Equation \ of \ the \ line \ through \ (-2, 2) \ and \ slope, m_3:\\(y - y_3) =m_3(x - x_3)\\\\(y-2) = \frac{1}{4} (x - (-2))\\\\y-2 = \frac{1}{4}(x+2)\\\\y = \frac{1}{4}x + \frac{1}{2} +2\\\\y = \frac{1}{4}x + \frac{5}{2}

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If ∠C and ∠D are complementary angles, m∠C = (4x + 3)°, find m∠D = (15x - 8)°, find m∠D.
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Answer:

67

Step-by-step explanation:

4x+3+15x-8=90(complementary angle)

x=5

now D = 15*5-8

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