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Rina8888 [55]
3 years ago
8

density is the ratio of mass to volume density = mass/volume compare the density of different liquids with the same volume the l

iquid with the greatest density is ____ the liquid wity the least density is ____
Mathematics
1 answer:
bija089 [108]3 years ago
7 0

Here, we are required to compare the density of different liquids with the same volume

Since density is the ratio of mass to volume, different liquids for example, kerosene and water if they have the same volume, will have different masses.

This is due to the difference in their densities.

Therefore, according to the statement in the question above;

While comparing the density of Liquids with the same volume, the liquid with the greatest density is heavier(more quantity of matter per unit volume) while the liquid with the least density is lighter( less quantity of matter per unit volume).

Read more:

brainly.com/question/17826958

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Can someone help me please
mojhsa [17]

Answer:

Your answer is B

Step-by-step explanation:

There are 3 kids ages 9-10, so that's the first 3.

There are 10 kids ages 10-11, so that's the second 10.

There are 6 kids ages 11-12, so that's the third 6.

add those kids up and you get 19

3 0
3 years ago
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What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
3 years ago
A group of 2 adults and 4 children pay $95 for admission to a water park. A
il63 [147K]

Answer:

2a + 4c = 95

3a +7c =155

Step-by-step explanation:

2a + 4c = 95

3a +7c =155

8 0
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Now let’s practice what you have learned.<br><br><br> Solve the equation.<br> 4y – 4 = 28
laiz [17]

Answer:

<h2>y=8</h2>

Step-by-step explanation:

4y-4=28

4y=28+4

4y=32

y=32/4

y=8

now plug in 8 for y and check the equation true:

4y-4=28

4*(8)-4=28

32-4=28

28=28

So,  y is equal to eight 8

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3 years ago
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KiRa [710]
You will pay $2.78 per mile for a total of 7 miles you will pay $19.50, but you must remember the taxi driver charges a $2.00 fixed rate so your final total will be $21.50
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