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Oksi-84 [34.3K]

3 years ago

15

9, 21, 7, 11, 13, 21, 5, 14, whats the range?

2 answers:

maria [59]3 years ago

4 0

16 would be the range. The lowest value is 5, and the highest value is 21. You just subtract those two which gives you 16! (:

Tanzania [10]3 years ago

4 0

Hello there Angieevans425!
Your Question(s):
$9, 21, 7, 11, 13, 21, 5, 14,$
Whats the range?
My Answer(s):
First, we need to figure out what range is. The range is <span>The difference between the lowest and highest values. So, we need to order the numbers from least to greatest. So order the numbers as:
</span>5, 7, 9, 11, 13, 14, 21, 21.
Then subtract the lowest number from the biggest number:
$21 - 5 = 16$
So, 16 is your range!
Hope I helped,
♥ Sir Chub-Chubs.

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alexira [117]

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<h2><em><u>4</u></em></h2>

Step-by-step explanation:

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3 years ago

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

kodGreya [7K]

Answer:

$sin(t) =\frac{\sqrt{7}}{4}$

Step-by-step explanation:

By definition we know that

$sec(t) = \frac{1}{cos(t)}$

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$cos ^ 2(t) = 1-sin ^ 2(t)$

As $sec(t) = -\frac{4}{3}$

Then

$sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}$

Now square both sides of the equation:

$cos^2(t) = (-\frac{3}{4})^2$

$cos^2(t) = \frac{9}{16}\\\\$

$1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}$

In the second quadrant sin (t) is positive. Then we take the positive root

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3 years ago

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SpyIntel [72]

Answer:

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Step-by-step explanation:

We can write the equation of a circle as

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(x--3)^2 + (y-4)^2 = 8^2

(x+3)^2 + (y-4)^2 = 64

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4 years ago