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3 years ago

14

HELP.... square root of y5

1 answer:

dmitriy555 [2]3 years ago

4 0

Sqrt (y^4 * y)
Sqrt y*4 = y^2
y^2 sqrt y

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Find the midpoint of the segment below and enter its coordinates as an

o-na [289]

Answer:

(2,5)

Step-by-step explanation:

(X<em>m</em>, Y<em>m</em>)= ( (x1+x2)/2, (y1+y2)/2 )

= (5 + (-1))/ 2, (5+5)/2

= (4)/2, (10)/2

= 2, 5

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2 years ago

If the conversion rate for Swiss francs is 1.68 francs = 1 U.S. dollar, then how many Swiss francs can be exchanged for $79.00 i

Likurg_2 [28]

If one U.S. dollar is as much as 1.68 Swiss francs, then 79,00 U.S. dollars will be as much as 79,00 * 1,68, which is 132,72 Swiss francs.

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3 years ago

Please help !!!!!!!!!!!!!!!!!!!<br><br><br><br>what is A

algol13

Answer:

Step-by-step explanation:

a=11/12

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2 years ago

The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig

natali 33 [55]

Answer:

a) the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

b) $y = 28.36^0 \ C$ ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

$\mathbf{y = A sin (Bt -C)+D}$

where A = amplitude and which can be determined via the formula:

$A = \dfrac{largest \ temperature - lowest \ temperature}{2}$

$A = \dfrac{29-14}{2}$

$A = \dfrac{15}{2}$

A = 7.5° C

where B = the frequency;

Since the data covers a period of 3 days ; then $\dfrac{2 \pi}{B } =365$

$B = \dfrac{2 \pi}{365}$ ( where 365 is the time period )

The vertical shift is found by the equation D;

D = $\frac{largest \ temperature + lowest \ temperature}{2}$

D = $\frac{29+14}{2}$

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

$y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5$

From the question; when it is 7th of the year ( i.e January 7);

t = 7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

$29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5$

$29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5$

$\frac{29-21.5}{7.5}= sin (\frac{14 \pi}{365} -C)$

$1= sin (\frac{14 \pi}{365} -C)$

$sin^{-1}(1)= (\frac{14 \pi}{365} -C)$

$\frac{\pi}{2}= (\frac{14 \pi}{365} -C)$

$C= (\frac{14 \pi}{365} -\frac{\pi}{2})$

$C= (\frac{28 \pi- 365 \pi}{730} )$

$C= \frac{-337 \pi}{730}$

Thus; the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

$y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5$

$y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5$

$y = 7.5 *( 0.915)+ 21.5$

$y = 6.8689+ 21.5$

$y = 28.36^0 \ C$ ( to two decimal places)

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Answer: A

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