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Advocard [28]
1 year ago
11

Two sides of an acute triangle measure 5 inches and 8 inches. The length of the longest side is unknown/ What is the greatest po

ssible whole-number length of the unknown side?
8 inches
9 inches
12 inches
13 inches
Mathematics
2 answers:
Nesterboy [21]1 year ago
7 0

Answer:

12in

Step-by-step explanation:

skelet666 [1.2K]1 year ago
3 0

The Triangle inequality theorem of a triangle says that the sum of any of the two sides of a triangle is always greater than the third side. The correct option is C.

<h3>What is the triangle inequality theorem?</h3>

The Triangle inequality theorem of a triangle says that the sum of any of the two sides of a triangle is always greater than the third side.

Suppose a, b and c are the three sides of a triangle. Thus according to this theorem,

 (a+b) > c\\(b+c) > a\\(c+a) > b

Given the length of the two sides of the triangle, therefore, we can write the inequalities,

8 + 5 > x  ⇒   13 > x

8 + x > 5  ⇒  x > - 3

5 + x > 8  ⇒  x > 3

Now, as per the inequality the value of x can lie between 3 to 13, but as the side needs to be greatest, therefore, the value of x will be 12.

Hence, the correct option is C.

Learn more about the Triangle Inequality Theorem:

brainly.com/question/342881

#SPJ1

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nikdorinn [45]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Problem 1:</u>

A standard deck of cards contains 52 cards, consisting of 13 spades. If you select only one randomly, the probability of that occurring would be 13/52 or 1/4. Since there are only 26 red cards in a standard deck, then the probability of selecting a red card would be 26/52 or 1/2. Because the two events are independent of each other, their probabilities are multiplied. Therefore, the probability of selecting a spade, and then replacing it in hopes of drawing a red card is (1/2)(1/4) = 1/8.

<u>Problem 2:</u>

We are selecting a spade and then another spade while NOT replacing the first spade (remember that these events are independent of each other also). This means that the total card count will change by picking up the second card. Therefore, the probability of selecting a spade, followed by another spade, is (13/52)(12/51) = 156/2652 = 1/17.

4 0
3 years ago
Find the expected value of the winnings
Orlov [11]

The expected value of the winnings from the game is $4

<h3>How to determine the expected value?</h3>

The payout probability distribution is given as:

Payout ($) 2 4 6 8 10

Probability 0.5 0.2 0.15 0.1 0.05

The expected value is then calculated as:

E(x) = \sum x * P(x)

This gives

E(x) = 2 * 0.5 + 4 * 0.2 + 6 * 0.15 + 8 * 0.1 + 10 * 0.05

Evaluate the expression

E(x) = 4

Hence, the expected value of the winnings from the game is $4

Read more about expected values at:

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7 0
1 year ago
PLEASE HELP ME! i think its A
natali 33 [55]

Answer:

The answer is 3rd option.

Step-by-step explanation:

In a graph, domain is always represented by the x-axis. So by looking at the graph, the minimum value of x is -2. We can know that x ≥ -2.

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3 years ago
PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS
Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

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

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0
1 year ago
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