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DIA [1.3K]
3 years ago
8

Please helpp meeee!!!

Mathematics
1 answer:
Law Incorporation [45]3 years ago
5 0

Answer:

im pretty sure the answer would be,

2 \frac{1}{2}

Step-by-step explanation:

the median is the middle number of the data, and since there are two medians, you find the mean of the two numbers by adding them together, and then you divide it by 2.

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A basketball player made 25 baskets in a season. Of these, 36% were three-point shots. How many three-point shots did he make? A
denis-greek [22]
<span>36% of 25 baskets were three-point shots
 = 0.36 x 25
= 9

So the correct answer is <u>D. 9</u></span>
4 0
3 years ago
45% of 40 is 18% of what number?
alexandr1967 [171]
So 45% OF 40 is 
45% * 40 
0.45 * 40
that is 18

18 is 18% of what number?
18 = 18% * x
18 = 0.18 * x
divide both sides by 0.18
100 = x
45 % of 40, which is 18, and is 18% of 100.

Hope this helps
3 0
3 years ago
Read 2 more answers
Lim x-&gt; vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))
NNADVOKAT [17]

I believe the given limit is

\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)

Let

a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1

Now rewrite the expression as a difference of cubes:

a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}

Then

a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2

The limit is then equivalent to

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}

From each remaining cube root expression, remove the cubic terms:

a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}

(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}

b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}

=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}

8 0
3 years ago
An ultramarathon relay of 37.2
Naya [18.7K]

Equation represent situation is 6d = 37.2

<u>Given:</u>

Total distance cover by runners = 37.2 miles

Number of runners = 6

<u>Find:</u>

Equation represent situation

<u>Computation:</u>

Assume;

Each runner cover distance = d miles

So.

Total distance cover by runners = Number of runners × Each runner cover distance

37.2 = 6 × d

6d = 37.2

Learn more:

brainly.com/question/15172156?referrer=searchResults

6 0
3 years ago
Help me plsssss I have more but this one first
Alexeev081 [22]

Answer:

1205.76   \:  \: {units}^{3}

Step-by-step explanation:

v = \pi {r}^{2} h \\  = 3.14 \times  {8}^{2}  \times 6 \\  = 3.14 \times 64 \times 6 \\  = 1205.76   \:  \: {units}^{3}

Hope this helps you

<em>Can</em><em> </em><em>I</em><em> </em><em>have</em><em> </em><em>the</em><em> </em><em>brainliest</em><em> </em><em>please</em><em>?</em>

4 0
3 years ago
Read 2 more answers
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