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Alborosie
3 years ago
12

The 6 members of the homecoming decorating committee want to make 525 paper flowers for the homecoming dance. Each flower takes

about 4 1/2 minutes to make. About how long will each member of the committee have to spend making flowers?
Mathematics
1 answer:
stiv31 [10]3 years ago
4 0
<h3>Answer:</h3>

6 hours 36 minutes, or 6 hours 31 1/2 minutes

<h3>Step-by-step explanation:</h3>

Dividing the task among the 6 committee members means each member will be making 525/6 = 87.5, that is, either 87 or 88 flowers. (We doubt that two people working together on the same flower will finish it in half the time.)

Each of the three committee members making 88 flowers will take ...

... 88 × 4 1/2 minutes = 396 minutes = 6 hours 36 minutes

Each member making 87 flowers will be finished 4 1/2 minutes sooner, after 6 hours 31 1/2 minutes.

_____

<em>Comment on the problem</em>

This problem seemingly invites you to divide the labor evenly among committe members. Doing that would give you an answer of 6 hours 33 3/4 minutes. You need to ask yourself whether that is practical for this situation.

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What is the distance between (7, 1) and (4, -5)?
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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.

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Sergio039 [100]

Answer:

  1. A = 2x; P = 4x+2. A = 4; P = 10.

  2. A = y² +2; P = 4y +2. A = 27; P = 22.

Step-by-step explanation:

1. The area is the sum of the marked areas of each of the tiles:

  A = x + x

  A = 2x

__

The perimeter is the sum of the outside edge dimensions of the tiles. Working clockwise from the upper left corner, the sum of exposed edge lengths is ...

  P = 1 + (x-1) + x + 1 + (x+1) + x

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__

When x=2, these values become ...

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_____

2. Again, the area is the sum of the marked areas:

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  A = y² +2

__

The edge dimension of the square y² tile is presumed to be y, so the perimeter (starting from upper left) is ...

  P = y +(y-2) +1 +2 +(y+1) +y

  P = 4y +2

__

When y=5, these values become ...

  A = 5² +2 = 27 . . . . square units

  P = 4·5 +2 = 22 . . . . units

4 0
4 years ago
If m∠4 = 70°, what is m∠6?
Karolina [17]

Answer:

6 i think

Step-by-step explanation:

7 0
3 years ago
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