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Goryan [66]
3 years ago
14

Marissa notices that the top of the net is 20 inches higher than the tallest player on the team. The net is less than 90 inches

high. She says that the inequality 20 + h < 90, where h is the height of the tallest player, shows this relationship. Is this correct? Explain why or why not.
Mathematics
2 answers:
Vika [28.1K]3 years ago
6 0
No, the inequality uses less than or equal to, but it should be less than. The relationship could be written as h<span> + 20 < 90. This says that the tallest player’s height plus 20 inches is less than 90 inches.</span>
hjlf3 years ago
6 0

Answer:

The inequality sign is not correct.

The height plus 20 needs to be less than 90.

Step-by-step explanation:

You just gotta do those two sentences. np :)

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Answer:

1. 8/$12=21/$x then cross multiply.

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3. $31.5

Step-by-step explanation:

12x 21/ 8 = 31.5

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Answer:

Step-by-step explanation:

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3 years ago
18 POINTS will rate brainiest 6th grade math
stira [4]

Answer:

18 ft^2

Step-by-step explanation:

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Can someone help me with this question please. I'll give the brainliest answer to whoever helps me.
Doss [256]

Answer: \sqrt[5]{y}

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and the more general rule we could use is

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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)

7 0
3 years ago
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