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Genrish500 [490]
3 years ago
11

The pie chart below shows the percentage of total revenue that a publisher receives from various types of publications. Use this

chart to answer the questions
below.

Mathematics
1 answer:
liberstina [14]3 years ago
7 0

Answer:

a. Cookbooks

b. 50%

c. 30%

Step-by-step explanation:

According To the Question,

  • We have a total revenue of a publisher describe in a circle (360°).

a. Now, Approximately Cookbooks & Textbooks revenue Form 180°, but textbook revenue is more than cookbooks as clearly visible in the diagram

Thus, cookbooks are less than 90°.

Now, we have to find 1/5th of total publisher revenue which is 360°/5= 72°. & the Cookbooks is nearest to 72° ( less than 90°)

So, Answer For (a) is Cookbooks  

b. Here in Diagram Clearly Visible that The Cookbooks & Textbooks Revenue Form 180° which is Approximately 50% of total Publisher's Revenue.

So, the Total Revenue Comes From Textbooks & Cookbooks is 50%.

c. Now We know the Cookbooks + Textbooks Revenue form 180° Approximately & Cookbook is approximately equal to 72° (as we solve above)

So, textbooks are 180°-72° = 108°, which is 30% of 360° ∴ 30% Revenue Come From Textbooks.

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Greg and Erin start at the same point and begin running in different directions. Greg is running east at a speed of 7 miles per
AveGali [126]

Answer: The time at which the distance between the two is 150 miles is 1.3 hr.

Step-by-step explanation:

Step 1: Sketch the problem as shown in the attached image.

The problem can be solved using Pythagoras theorem since the sketch is a right angled triangle.

It follows the equation x² + y² = 15².

The distance covered by Greg is given by <em>x mi</em>, and that of Erin is given by <em>y mi</em>.

Step 2: To derive the distance covered by Erin and Greg, we use the formula <em>Speed = distance ÷ time.</em>

Which by cross-multiplying gives <em>distance = speed × time</em>.

Let time taken to cover the distance be <em>t h</em>.

Given that Speed of Erin is 9 mph and that of Greg is 7 mph.

Hence, distance covered by Greg, x = 7 × t = 7t.

Distance covered by Erin, y = 9 × t = 9t.

Step 3: Insert the terms x and y into the Pythagoras equation x² + y² = 15² and solve for <em>time t.</em>

(7t)² + (9t)² = 15²

49t² + 81t² = 225

(49 + 81)t² = 225

      t² = 225/(49 + 81)

      t² = 225/130

      t² = 1.730769231

       t = √1.730769231

       t = 1.315587029

t = 1.3hr

Hence the time at which the distance between the two is 150 miles is 1.3 hr.

4 0
3 years ago
The radius of a truck wheel is 3 ft. what is the approximate circumference if the truck wheel.
Phoenix [80]
Circumference = 2 \pi r
When you plug that in you should have (2* \pi ) *3
Solve that and you have your answer(: I would recommend rounding to the nearest hundreth
8 0
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Answer:

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4 0
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Can a quadrilateral be a rhombus ?
koban [17]

Yes it can.  It doesn't <em>have to be</em>, but it can be without too much trouble.

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6 0
3 years ago
Change the expression to a single square root, or its opposite:
aleksley [76]

Answer:

a)2\sqrt{2}=\sqrt{8}

b)-7\sqrt{3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b}  =\sqrt{2.b}

d)5\sqrt{y} =\sqrt{25y}

e)-6\sqrt{2a}  =-\sqrt{72a}

f)-0.1\sqrt{200c}=-  \sqrt{2c

Step-by-step explanation:

a) 2\sqrt{2}=( \sqrt{2}  )^2.\sqrt{2}=(\sqrt{2})^3 = \sqrt{2^3} =\sqrt{8}

b)-7\sqrt{3} =-(\sqrt{7} )^2\sqrt{3} =-\sqrt{7^2.3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b} =\frac{1}{3} \sqrt{9.2.b} =\frac{1}{3} \sqrt{3^2.2.b} =\frac{1}{3} \times 3\sqrt{2.b} =\sqrt{2.b}

d)5\sqrt{y} =\sqrt{5^2} \sqrt{y}=\sqrt{25y}

e)-6\sqrt{2a} =-\sqrt{6^2}\sqrt{2a}  = -\sqrt{36.2a} =-\sqrt{72a}

f)-0.1\sqrt{200c}=-\frac{1}{10}  \sqrt{10^2.2c} =-\frac{1}{10}\times10  \sqrt{2c}=-  \sqrt{2c

4 0
3 years ago
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