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Harman [31]
3 years ago
5

The picture shows a feeding trough that is shaped like a right prism. (Picture is shown at the bottom!!)

Mathematics
1 answer:
AlexFokin [52]3 years ago
4 0
Height of the trapezoids = sqrt (17^2 - 8^2) = 15cms
base = 19 cms

(b) area of each trapezoid =  15/2 (35 + 19) =  405 cm^2

Total area painted blue = 2*405 = 810 cm^2

(c)  length of the edges of one trapezoid = 35 + 19 + 2(17) = 88 cms
So one piece of  sandpaper ( can sand up to 80 cm) will not be enough to sand all of the edges.
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Please tell me the final answer to this question and the process. Thanks!
shusha [124]
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The amount of time it takes a swimmer to swim a race varies inversely as the average speed of the swimmer. A swimmer finishes a
yarga [219]

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5 average speed

Step-by-step explanation:

3 0
3 years ago
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HELP PLEASE WILL MARK BRAINLIEST!!
Arte-miy333 [17]

Answer:

Answer is option d)346.4 ft

Step-by-step explanation:

from \: the \: figure \\ AB = 300 \: ft \:  \:  \: angle \: C = 60  \:( theta)\: degress \\ in \: triangle \: ABC \: using \:  tan \: theta\\  \tan(60)  =  \frac{opposite \: side}{adjacent \: side}  \\  \tan(60)  =  \frac{300}{BC}  \\  \sqrt{3}  =  \frac{300}{BC}  \\ BC =  \frac{300}{ \sqrt{3} }  \\ BC = 100 \sqrt{3} ft \\ then \: in \: triangle \: ABD \: using \: tan \: theta \\ angle \: D = 30 \: degrees \: (theta) \\   \tan(30 )  =\frac{opposite \: side}{adjacent \: side} \\  \tan(30)   =  \frac{300}{BD}  \\  \frac{1}{ \sqrt{ 3} }  =  \frac{300}{BD}  \\ BD = 300 \sqrt{3}  \\ CD = BD - BC \\  = 300 \sqrt{3 }  - 100 \sqrt{3}  \\  = (300 - 100) \sqrt{3}  \\  = 200 \sqrt{3 }  \\  = 200 \times 1.732 \\  = 346.4 \: ft

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

7 0
3 years ago
How do you solve x/3&gt;-1
Goryan [66]

x/3>-1

Isolate the x. Multiply 3 to both sides

(x/3)3 > -1(3)

x > -1(3)

x > -3

x > -3 is your answer

hope this helps

7 0
3 years ago
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3 years ago
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