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kogti [31]
2 years ago
14

What is the period of the graph of y = 5cos(6x – 1) + 3? O A.

Mathematics
1 answer:
evablogger [386]2 years ago
5 0

Step-by-step explanation:

The coefficent of the x variable is b.

The formula for the period is

\frac{2\pi}{ |b| }

Here, b is 6 so we have

\frac{2\pi}{ |6| }  =  \frac{\pi}{3}

So the period is pi/3, or 60 degrees

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The data shown has a least-squares regression line of y=3.2x−6.4.
Damm [24]

Answer:

1) B. on, 2) D. 19.2, 3) C. 0

Step-by-step explanation:

1) The value of y according to the regression line is:

y = 3.2\cdot (8) - 6.4

y = 19.2

Hence, the point (8,19.2) is <em>on </em>the least-squares regression line.

2) The value of y according to the regression line is 19.2.

3) The residual is the difference between the value from the regression line and the real value. In this case, the residual value is 0.

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3 years ago
Ethan was busying marshmallows for a family bonfire. Each bag of marshmallows was 250 g. If Ethan wanted to buy 3 kg of marshmal
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It's 12 bags (3000/250)
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7. Lakeisha is designing a rectangular sunroom. She wants the room to be 14 feet wide and have a perimeter that is greater than
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D and E is the answer. C is exactly 108 and not greater than

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In Las Vegas, the hottest recorded temperature is 47˚C, and the lowest recorded temperature is -13˚C. What is the difference bet
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Through (-1,4) and a perpendicular to 4y-2x=12
mart [117]

Answer:

<h2>y = -2x + 2 → 2x + y = 2</h2>

Step-by-step explanation:

\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-\text{slope}\\b-\text{y-intercept}\\\\\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\\text{then}\\\\k\ \perp\ l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\k\ ||\ l\iff m_1=m_2

\text{We have the equation of a line in the standard form:}\ Ax+By=C.\\\\\text{Convert to the slope-intercept form:}\\\\4y-2x=12\qquad\text{add}\ 2x\ \text{to the both sides}\\\\4y-2x+2x=2x+12\\\\4y=2x+12\qquad\text{divide both sides by 4}\\\\\dfrac{4y}{4}=\dfrac{2x}{4}+\dfrac{12}{4}\\\\y=\dfrac{1}{2}x+3\to\boxed{m_1=\dfrac{1}{2}}

\text{Therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2.\\\\\text{Put the value of the slope and the coordinates of the given point (-1, 4)}\\\text{to the equation of a line:}\\\\4=-2(-1)+b\\\\4=2+b\qquad\text{subtract 2 from the both sides}\\\\4-2=2-2+b\\\\2=b\to b=2\\\\\bold{FINALLY:}\ y=-2x+2

\text{Convert to the standard form:}\\\\y=-2x+2\qquad\text{add}\ 2x\ \text{to the both sides}\\\\y+2x=-2x+2x+2\\\\2x+y=2

4 0
4 years ago
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