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4 years ago

Solve the inequaliy: (x^2+2)(x−2)^2(6−2x)>0

Mathematics

1 answer:

marin [14]4 years ago

7 0

The answer is x<2
-2x^5 + 14x^4 - 36x^3 + 52x^2 - 64x + 48=0

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Select all the expressions that equal.. *see picture for problem*

kupik [55]

Answer:

Step-by-step explanation:

oknjbgfkdfrgn jnfbfl

7 0

3 years ago

Can someone help me with these

Lapatulllka [165]

Is this due tmr or today

5 0

3 years ago

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

4 years ago

Write a variable expression to describe the rule for the sequence. Then find the 100th term. 35, 36, 37,...

Dmitry [639]

Given expression is 35, 36, 37, 38...

First term a = 35

Common difference d = 36-35 = 1

Nth term of arithmetic sequence is:

$n_{th} = a+ (n-1)d$

$n_{th}= 35+(n-1)(1)$

$n_{th} = 34+n$

To find the 100th term, substitute n = 100

$n_{100} = 34 +100$

$n_{100} =134$

Correct option is B

5 0

3 years ago

Determine a series of transformations that would map polygon ABCDE onto polygon A’B’C’D’E

zalisa [80]

Answer:

Reflected across x-axis followed by a dilation with a scale factor of 2 about the origin.

Step-by-step explanation:

It's clear from the graph,

Polygon ABCDE is reflected across x-axis then dilated so that it overlaps polygon A'B'C'D'E'

Let's choose a point A having coordinates (-2, -1)

By reflecting point A across x-axis,

Rule to be followed for the reflection,

(x, y) → (x, -y)

By this rule, coordinates of the image point A will be,

A(-2, -1) → A"(-2, 1)

Now point A"(-2, 1) has been dilated by a scale factor = k to form an image points A'(-4, 2).

Rule for the dilation of a point about the origin is,

A"(x, y) → A'(kx, ky)

A"(-2, 1) → A'(-2k, k)

Since, coordinates of point A' are (-4, -2),

Therefore, -2k = -4

k = 2

Hence, polygon ABCDE was reflected across x-axis followed by a dilation with a scale factor of 2 about the origin to form polygon A'B'C'D'E'.

3 0

3 years ago