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Musya8 [376]
2 years ago
13

The following set of ordered pairs represents a power inverse variation. Find the value of r given that k = 5.

Mathematics
1 answer:
pochemuha2 years ago
4 0

Answer:

r=2

Step-by-step explanation:

y = k x^r  is the formula for a direct variation

y = k x^ -r is the formula for a indirect variation

20= 5 (1/2)^ -r

Divide each side by 5

4 = (1/2) ^ -r

Rewriting

2^2 = 2^ -1 ^ -r

2^2 = 2 ^ r

The bases are the same so

2 =r

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5x-2=3x+8


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That's your answer
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A savings account earns 5% simple interest per year. The principal is $1200. What is the balance after 4 years?
stiv31 [10]
The formula of the Simple Interest is:
I=PRT
P for Principle Amount     ($1200)
R for Rate                        (5%=\frac{5}{100} = 0.05)
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I = 1200 × 0.05 × 4
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Add the interest to the principle amount to check the balance
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3 years ago
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In a class of 55 students, 15 students liked Maths but not English and 18 students liked English but
ra1l [238]
Easy!
You take the total number of students (55) then you remove the people that only like math and remove the people that only like English and remove the people that like none of them and what’s left is how many like both subjects.

55-15-18-5=17 students that like both subjects!

6 0
2 years ago
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
2 years ago
A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 50 percent of its previous height. The total vert
xz_007 [3.2K]
After every drop,the ball bounces to half it's previous height. With that understood.

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2nd drop - 5m down

2nd bounce - 2.5m up
3rd drop - 2.5m down

3rd bounce - 1.25m up
4th drop - 1.25m down

4th bounce - 0.625m up
5th/last drop - 0.625m down

To find the total vertical distance, you add them all.

10+5+5+2.5+2.5+1.25+1.25+0.625+0.625
=29.25m travelled in all.
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3 years ago
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