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4 years ago

The variables x and y are inversely proportional, and y = 2 when x = 3. What is the value of y when x = 9?

2 answers:

7 0

$If\ x\ and\ y\ are\ inversely\ proportional\ then\ xy\ is\ constans.\\\\9y=2\times 3\\\\9y=6\ \ \ \ \ \ /:9\\\\y=\frac{6}{9}\\\\y=\frac{2}{3}$

kari74 [83]4 years ago

5 0

Y=k/x
2=k/3
k=6
y=6/x
y=6/9=2/3

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What is the slope of a trend line that passes through the points (5, 80) and (7, 65)?

Sedaia [141]

The slope of a trend line that passes through the points (5, 80) and (7, 65) is -7.5.

Step-by-step explanation:

If the line is passing through two points (x1, y1) and (X2, y2). The slope (m) can be calculated using following formula.

$m = \frac{y2-y1}{x2-x1}$

In this case,

x1 = 5

x2 = 7

y1 = 80

y2 = 65

By putting the values in formula, we get

$m = \frac{(65-80)}{7-5}$

$m = \frac{(-15)}{2}$

m = -7.5

Therefore, the slope of a trend line that passes through the points (5, 80) and (7, 65) is -7.5.

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4 years ago

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When you divide the product of n consecutive integers by a natural number K, the remainder is a positive number. Find the greate

tatyana61 [14]

It would be 34295829

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3 years ago

If 5x:(x-2) = 11:3 , find the value of x ?

Masteriza [31]

Answer:

x = -11/2

Step-by-step explanation:

The ratios need to stay the same

5x 11

----- = -------

x-2 3

Using cross products

5x*3 = (x-2) 11

15x = 11x - 22

Subtract 11 from each side

15x-11x = -22

4x = -22

x = -22/4

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3 years ago

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What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

jasenka [17]

<u>Answer:</u>

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

<u>Solution: </u>

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$