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Eddi Din [679]
3 years ago
10

The ratio of the measures of the sides of a triangle is 5:7:8 and has a perimeter of 320 inches. What is the measure of each sid

e?
Mathematics
1 answer:
Readme [11.4K]3 years ago
8 0
<span>The ratio of the measures of the sides of a triangle is 5:7:8, and its perimeter is 40 inches. what is the measures of each side of the triangle 
Sides are (5/20) x 40 = 10 inches
(7/20) x 40 = 14 inches, and (8/20) x 40 = 16 inches respectively. </span>
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When looking at the graph of a 5th degree function, how can you determine if all of the zeros of the function are real?
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Answer:

If it cuts x-axis 5 times.

Step-by-step explanation:

When we look at the graph of a function we can see its real roots by looking at its graph

The intersecting points that is the number of times a line cutting x-axis will be the real root of the function

So, by looking at the 5th degree function the number of time that function cuts x-axis will be the number of real roots.

So, if we need to say all the zeroes or roots of the function are real means it will cut the x-axis 5 times.

Because a function will have the root equal to its degree.

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Exercise 10.10.2: Distributing a coin collection. About A man is distributing his coin collection with 35 coins to his five gran
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

Given that:

all coins are same;

The same implies that the number of the non-negative integral solution of the equation:

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

Thus, the number of the non-negative integral solution is:

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(b)

Here all coins are distinct.

So; the number of distribution appears to be an equal number of ways in arranging 35 different objects as well as 5 - 1 - 4 identical objects

i.e.

= \dfrac{(35+4)!}{4!}

= \dfrac{39!}{4!}

(c)

Here; provided that the coins are the same and each grandchild gets the same.

Then;

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

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5x_1 = 35\\\\ x_1= \dfrac{35}{5} \\ \\  x_1= 7

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(d)

Here; we need to divide the 35 coins into 5 groups, this process will be followed by distributing the coin.

The number of ways to group them into 5 groups = \dfrac{35!}{(7!)^55!}

Now, distributing them, we have:

\mathbf{\dfrac{35!}{(7!)^55!}  \times 5!= \dfrac{35!}{(7!)^5}}

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