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nadezda [96]
2 years ago
7

The table below contains four statements.

Mathematics
2 answers:
alexandr1967 [171]2 years ago
4 0

Answer:

  f(x)

Step-by-step explanation:

The value of y will be zero at the x-intercept. Using x=27 in the four functions, we find ...

  k(27) = ∛(-54) ≠ 0

  h(27) = 3∛0 = 0

  g(27) = 3 +∛0 = 3

  f(27) = ∛0 = 0

So, choices h(x) and f(x) match the x-intercept requirement.

__

When x=0, the function value is the y-intercept.

  h(0) = 3∛27 = 3·3 = 9

  f(0) = ∛27 = 3

The only function to match the x- and y-intercept requirements is f(x). It also matches the requirements for the domain and the slope.

kirill115 [55]2 years ago
3 0

Answer:

Hey your answer is d I used this app called desmos graphing calculator it’s very helpful for graphs. Hope I helped!!

Step-by-step explanation:

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Simplify the polynomial. Then find its value for the given value of the variable.
vekshin1

Answer:

Step-by-step explanation:

Simplified expression:

3x^3 -x^3 +5x^3-9x^3

2x^3 +5x^3-9x^3

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If x = -1, the value of the expression is:

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3 0
3 years ago
Find (a) the number of subsets and (b) the number of proper subsets of the set.
vaieri [72.5K]

Answer:

(a) Total No. of Subsets = 128

(b) Total No. of Proper Subsets = 127

Step-by-step explanation:

First we need to define the set of days of the week.

Set of Days of Week = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

It is evident from the set of days of the week, that it contains 7 elements.

(a)

The total no. of subsets of a given set is given by the following formula:

Total No. of Subsets = 2^n

where,

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Therefore,

Total No. of Subsets = 2^n

Total No. of Subsets = 2^7

<u>Total No. of Subsets = 128</u>

(b)

The total no. of proper subsets of a given set is given by the following formula:

Total No. of Proper Subsets = (2^n) - 1

where,

n = no. of elements of the set = 7

Therefore,

Total No. of Proper Subsets = (2^n) - 1

Total No. of Proper Subsets = (2^7) - 1 = 128 - 1

<u>Total No. of Proper Subsets = 127 </u>

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3 years ago
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