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Vinvika [58]
2 years ago
15

Condense log₂ 4 + log₂ 5

Mathematics
2 answers:
yawa3891 [41]2 years ago
7 0

Answer:   \log_2(20)

Work Shown:

\log_2(4) + \log_2(5)\\\\\log_2(4*5)\\\\\log_2(20)\\\\

The rule used is \log_{b}(\text{x}) + \log_{b}(\text{y}) = \log_{b}(\text{xy})

liberstina [14]2 years ago
3 0

\large{\textbf{Heya !}}

✏\large\sf{\bigstar Given:-} ✏

  • ㏒ expression = \sf{\log_24+\log_25}

✏\large{\sf{\bigstar To\quad Find:-} ✏

  • Condense the ㏒ into ONE expression - ?

✏\large{\sf{\bigstar Solution\quad steps:-}

<u></u>

<u>`To find what we want we need to apply ㏒ rules </u>

<u />

<u />

<u />

<u />\large{\sf{\longmapsto{log_ba+log_bc=log_b(ac)}

<u>using formula,</u>

<u />

<u />

<u />

<u />\large{\sf{\longmapsto{log_24+log_25=log_2(4\cdot5)}

<u>simplifying,</u>

<u />

<u />\large{\sf{\longmapsto{log_220}

<u>`hope it was helpful ! ~</u>

<u />

<u />

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A television transmitter tower is 400 feet high. if the angle between the guy wire (attached at the top) and the tower is 51.8°,
vova2212 [387]
Let the length of the guy wire be x feet.
cos 51.8 = 400/x
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The answer is 646.82 feet.
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 An athlete runs an equal distance 2 days a week. The other 5 days of the​ week, he runs a total of 11 miles. Write an equation
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Ok since the athlete an equal distance of 2 days a week and he runs a total of 11 miles in 5 days you will need to think back so before he had 11 miles done next weak he had a total of 19 miles so you will start at 11 then count up to 19.

Use a number line to show how many days it took for the athlete to get 19 miles.

Then once you do the number line count the lumps and that’s how many days it took for the athlete to get 19 miles.

Hope this helped.
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What is the formula for slope-intercept form?
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3 years ago
1. what is the period of the function f(x) shown in the graph?
ki77a [65]
Problem 1)

Answer: pi/2

-------------------------------

Place your pencil at (0,3) which is a good starting point, since x = 0 is often a good starting point. Trace along the curve until you reach back up at y = 3 again. So you'll go down and then back up until you reach (pi/2, 3). The difference in x values is pi/2-0 = pi/2. Every pi/2 units, the graph curve repeats itself over and over infinitely. 

Note: this is not the only way to determine the period. You can start at any point really. It helps to start at a min or max value, or at x = 0. In this case, it happened to be both. 

===========================================================
Problem 2)

Answer: See figure 2 (attached image)

-------------------------------

Point A is at (0,0)
Point B is at approximately (1.57,-2) which is exactly (pi/2,-2)

===========================================================
Problem 3)

Answer: See figure 3 (attached image)

===========================================================
Problem 4)

Answer: Choice B
The graph is horizontally compressed by a factor of 2 and shifts up 1 unit

-------------------------------

The jump from x to 2x will have the period go from 2pi to pi. The graph curve repeats itself more and more often (twice as often). Therefore we have an accordion squeeze effect going on, or think of it like a spring being pressed down. We have horizontal compression along the x axis.

Adding 1 to the end shifts every point upward 1 unit. Overall, this gives the visual the entire curve itself is shifted up 1 unit.

===========================================================
Problem 5)

Answer: f(x) = 5.05*sin((pi/12)x)+5.15

-------------------------------

max = 10.2
min = 0.1
difference = 10.2-0.1 = 10.1
Divide this difference in half: 10.1 = 5.05
So a = 5.05 is the amplitude

The period is T = 24 hrs since the height is 5.15 at midnight of one day, it rises and then falls and rises back to 5.15 by the next midnight
b = 2pi/T = 2pi/24 = pi/12
c = 0 since there is no phase shift in this sine function

d = 5.15 is the midline, which is the starting height at x = 0

f(x) = a*sin(bx-c)+d
f(x) = 5.05*sin((pi/12)x-0)+5.15
f(x) = 5.05*sin((pi/12)x)+5.15

===========================================================
Problem 6)

Answers are:
A) The max height of the Ferris wheel is 64 ft
B) The radius of the Ferris wheel is 30 ft

-------------------------------

The largest height of the table is 64, so 64 is the max height. This is why A is true.

B is true since the max is 64 and the min is 4, so 64-4 = 60 is the diameter, making the radius r = d/2 = 60/2 = 30

C is false because the height at t = 0 is 34 ft, but the lowest point is at height h = 4

D is false because the period is actually 10 seconds. Eg: going from t = 7.5 to t = 17.5 is a time of 10 seconds. During this time, the heights are 4, 34, 64, 34, 4 telling us that a full cycle has taken place.

===========================================================

4 0
3 years ago
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