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Reptile [31]
1 year ago
6

Find: (4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y)

Mathematics
2 answers:
const2013 [10]1 year ago
7 0

Answer:

The coefficients are

11,-4 and 0

Step-by-step explanation:

We are given two algebraic expressions. We are subtracting the second from the first.

First one is

4x^2y^3+2xy^2-2y

Second expression is

-7x^2y^3+6xy^2-2y

The given expression

=4x^2y^3+2xy^2-2y-(-7x^2y^3+6xy^2-2y)\\=4x^2y^3+2xy^2-2y+7x^2y^3-6xy^2+2y\\=11x^2y^3-4xy^2+0y

The coefficients are

11,-4 and 0

MA_775_DIABLO [31]1 year ago
4 0

Answer:

11 -4 0

Step-by-step explanation:

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Answer:  2x + 15

5 0
2 years ago
Which term in the arithmetic sequence 11, 4,-3 . . . is -129?​
telo118 [61]

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Step-by-step explanation:

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4 0
2 years ago
Given f(x) and g(x) = f(k⋅x), use the graph to determine the value of k.
yan [13]

Answer:

The value of k is 3.

Step-by-step explanation:

The function f(x) passes through the points (0,4) and (-6,-2)

So the equation of the function is \frac{y-4}{4-(-2)} = \frac{x-0}{0-(-6)}

⇒ y = x + 4 ....... (1)

Again the function g(x) passes through the points (0,4) and (-2,-2).

Therefore, the equation of g(x) will be  \frac{y-4}{4-(-2)} =\frac{x-0}{0-(-2)}

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Therefore, g(x) = 3x + 4 = f(3x) {from equation (1).

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3 0
1 year ago
Read 2 more answers
A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
a box holds nine smaller boxes. each of the smaller boxes holds nine plastic containers. each plastic contai.er holds nine bags.
svp [43]
So, basically ..you just multiply
9*9*9*9
6561
4 0
2 years ago
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