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3 years ago

Each person is dealt 5 cards, show the total number of cards dealt for each players from 3 to 6 write the ratio of cards dealt

Mathematics

1 answer:

Fiesta28 [93]3 years ago

5 0

Answer:

Following are the responses to these question:

Step-by-step explanation:

Because every player has 5 cards handed

The card ratio is 5:1 per player.

So, if there are three teams

5 times 3=15 Cards

Four players=four times five=20 cards

Five players=5 times five=25 cards

6 cards=6 times 5=30 cards

The card ratio is 5:1 per player.

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For triangle ABC, find the measure of AB given me∠ A = 55 degrees, m∠b= 44 degrees, and b=6. . A. 45.22. B.96.68. C.88.19. .

jekas [21]

There exist a similar question where b = 68 instead of 6. First, determine the measure of the third angle, angle C,
m∠c = 180 - (55° + 44°) = 81°
Let x be the side AB, that which is opposite to angle C. Through the Sine Law,
68 / sin 44° = x / sin 81°
From the equation, the value of x is equal to 96.68. Thus, the answer is letter B.

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Anika [276]

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3 years ago

The city of Jasper has a water tower that holds 1,325,000 gallons of water. Express this number in scientific notation.

Tatiana [17]

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In ΔMNO, m = 780 inches, o = 760 inches and ∠O=164°. Find all possible values of ∠M, to the nearest degree.

Mademuasel [1]

Answer:

So the answer is both

180 and 328 (but it may or may not show you "not possible, if so, then you got it right)

Step-by-step explanation:

SinA/a= SinB/b

1. SinM/780 = Sin164/760

2. 780sin164/760 = 0.2828909704

3. M = sin^-1 (0.2828909704)= 16.4328229 or 16

Check for possibility

180-16= 164

164 + 16= 180 (not possible)

164 + 164= 328 (not possible)

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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