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3 years ago

Select the correct answer.

Mathematics

1 answer:

Law Incorporation [45]3 years ago

5 0

i think its b

Step-by-step explanation:

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A 10-ft vertical post casts a 14-inch shadow at the same time a nearby cell phone tower cast a 119-ft shadow. How tall is the ce

Yakvenalex [24]

Answer:

The cellphone towers height is 1020 ft.

Step-by-step explanation:

This question is solved using proportions, by a rule of three.

A 10-ft post casts a 14-inch shadow.

Each feet has 12 inches, which means that a post with 10 feet will cast a 14/12 = 1.167 ft shadow. How tall is the cell-phone tower with a 119-feet shadow?

10 feet post - 1.167 ft shadow

x feet tower - 119 feet shadow

Applying cross multiplication:

$1.167x = 10*119$

$x = \frac{1190}{1.167}$

$x = 1020$

The cellphone towers height is 1020 ft.

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3 years ago

Use completing the square to solve this quadratic equation x^2+10x+25=2

never [62]

answer:

( x + 5 )^2 = 2

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A campus radio station surveyed 500 students to determine the types of music they like. the survey revealed that 199 like rock,

Ostrovityanka [42]

<span>There are 24 who like both rock and country. There are 8 who like all 3 types, and these eight have been counted under the 24. This means that the number who like rock and country but not jazz is 24 - 8 = 16.
We are given a total of 155 who like country. We subtract the 16 who like both rock and country, and are left with 139 people who like country but not rock. (It does not matter whether they like jazz or not).
The probability is 155 out of the total of 500, so 155/500 = 31/100 = 0.31.
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If logarithm of 5832 be 6, Find thw base?

Lunna [17]

Answer:

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6 0

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