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hodyreva [135]
4 years ago
14

A circle with radius of 5cm sits inside a 11cm by 11cm rectangle

Mathematics
1 answer:
Sergeeva-Olga [200]4 years ago
5 0

OK, wonderful.  It should be very comfortable in there, since it's so small and fits so easily.

The circle can be placed completely inside the rectangle, in positions where it touches one side, two sides, or no sides.  It's too small to touch thee sides or four sides all at the same time.

Do you have some kind of question to ask regarding the situation ?

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The shark are fed three times a day during the morning feeding 2/15 of a ton. During the afternoon feeding the weight of fish fe
UNO [17]

Answer:

The shark are fed \frac16 \ ton of fish during the night.

Step-by-step explanation:

Given:

Weight of fish fed in the morning = \frac{2}{15}\ ton

Also Given:

During the afternoon feeding the weight of fish fed will be 1/15 of a ton more than the fish fed during the morning.

Weight of fish fed in the afternoon = \frac{2}{15}+\frac{1}{15}=\frac{2+1}{15}=\frac{3}{15}\ ton

Total fish fed in whole day = \frac12 \ ton

We need to find the Weight of fish fed in the night.

Solution:

Now we can say that;

Weight of fish fed in the night can be calculated by by subtracting Weight of fish fed in the morning and Weight of fish fed in the afternoon from Total fish fed in whole day .

framing in equation form we get;

Weight of fish fed in the night = \frac{1}{2}-\frac{2}{15}-\frac{3}{15}= \frac{1}{2}-(\frac{2}{15}+\frac{3}{15})=\frac{1}{2}-\frac{2+3}{15}= \frac{1}{2}-\frac{5}{15} = \frac{1}{2}-\frac{1}{3}

Now we will use LCM for making the denominator common we get;

Weight of fish fed in the night = \frac{1\times3}{2\times3}-\frac{1\times2}{3\times2} = \frac{3}{6}-\frac{2}{6}

Now denominator are common so we will solve the numerators.

Weight of fish fed in the night = \frac{3-2}{6}=\frac16 \ ton

Hence The shark are fed \frac16 \ ton of fish during the night.

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3 years ago
Solve for x cx + mx - y = z​
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Answer:

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Step-by-step explanation:

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4 years ago
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To two decimal places, find the value of k that will make the function f(x) continuous everywhere. 3x+k
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solution:

we are consider the following function,

f(x)=3x+k,x\leq 3

=kx^{2}-6,x>3

\lim_{x\rightarrow 3^-}(3x+k)=9+k

\lim_{x\rightarrow3^+}

(kx^{2}-6)=9k-6

so the left and right limits are equal.

therefore, the function is continuous at x=3

so,the therom of the function is continous at x=3

9k-6=9+k

8k=15

k=15/8

=1.875

therefore,the value of k=1.875

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One thousand and one hundred and thirty four divided by twenty seven
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Just use a calculator
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