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3 years ago

Ben can type 2/5 of a page of an essay in 1/2 minute. How much time would

Mathematics

2 answers:

KiRa [710]3 years ago

7 0

Answer:

ok so first i want to figure out what 2/5 converted to percent is......

you can find it out on a calculator or on paper by dividing 2 / 5 = 0.4

so this is 40% and if it takes him half a minute to do this, it takes him about 30 seconds to finish 40% of the essay and there is a full 60% still waiting to be done, so lets see,

40% + 40% = 80%

so that is about 1 full minute to write 80% we are still left with 20% so let's see how much time does it take to write 20% of the essay, in 40% it takes 30 seconds and in 80% it takes 60 seconds. So since 20% is half of 40% then we must divide the time in half so 30 / 2 = 15

So if we add all these numbers up it equals 1:15 minutes to write an essay

I hope this helps and please mark brainliest if i got it right!

DiKsa [7]3 years ago

4 0

Answer:

1.25 minutes, or 1 minute 15 seconds

Step-by-step explanation:

We know that is speed is 0.4 pages per 0.5 minutes. This is 0.8 pages per 1 minute. If we divide 1/0.8, we get 1.25 or 1 minute 15 seconds.

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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.

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3 years ago

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Maksim231197 [3]

Answer:

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3 years ago

What is the solution set for 2x+5y>-1 and 4x-3<-3?

bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

$\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}$

$\displaystyle\ x > \frac{-1 - 5y}{2}$

So, the solution for x in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 - 5y}{2}$

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

$\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}$

$\displaystyle\ x > \frac{-1 -1x}{5}$

So, the solution for y in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 -1x}{5}$

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

$\displaystyle\frac{4x}{4}$

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

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