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Nat2105 [25]
3 years ago
9

Give an example of a benchmark fraction and an example of a mixed fraction.

Mathematics
2 answers:
34kurt3 years ago
8 0

Answer:

Benchmark fractions: 4/12    5/12        6/12           7/12  

Mixed Fractions: 1 1/2      2 3/4               5 1/4          8 2/8

Step-by-step explanation:

A mixed fraction is simply a fraction with a whole number.

A benchmark fraction is a fraction used to compare and order other fractions.

prisoha [69]3 years ago
8 0

Answer:  Benchmark fractions are easy to visualize and identify, and thus, help in estimating the parts. When comparing two fractions with different numerators and denominators, we can either make their denominators common or compare them to a benchmark fraction such as 1/2.

First you will have 1 which is a whole then 1/2 + 1/2 =1 and so on and so forth.

Mixed fractions are like= 2 1/3 4 1/5 They are like whole numbers with fractions as what I mentioned before.

Hope you understand this :)

Step-by-step explanation:

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Find the volume of the cube below.
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The volume of the cube is 9
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3 years ago
The value 5 is an upper bound for the zeros of the function shown below. f(x)=x^4+x^3 -11x^2-9x+18
Inessa [10]
Answer: True

Explanation:
According to the rational zeros theorem, if x=a is a zero of the function f(x), then f(a) = 0.

Given:  f(x) = x⁴ + x³ - 11x² - 9x + 18
From the calculator, obtain
f(5) = 448
f(4) = 126
f(3) = 0
f(2) = -20
f(1) = 0
f(0) = 18
f(-1) = 16
f(-2) = 0
f(-3) = 0

The polynomial is of degree 4, so it has at most 4 real zeros.
From the calculations, we found all 4 zeros as x = -3, -2, 1, and 3.
Therefore
f(x) = (x+3)(x+2)(x-1)(x-3).

For x>3, f(x)increases rapidly. Therefore there are no zeros for x>3.
The statement that x=5 is an upper bound for the zeros of f(x) is true.



7 0
4 years ago
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sattari [20]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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#10 using the right angle below find the cosine of angle B.
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3 years ago
Find the complement of the set given that
iris [78.8K]

Answer:

A ' = {-2,-1,0,1, 4,7}

Step-by-step explanation:

Here, the Universal set U is given as:

U = {x | x is in I and −3 ≤ x ≤ 7}

⇒ U = { -3,-2,-1,0,1,2,3,4,5,6,7}

Now given: A = {−3, 2, 3, 5, 6}

Compliment  A set of any set A is the set of all the elements which is in the universal set U but NOT in A.

So, here the set of elements which are in U but not in A  is

A ' = {-2,-1,0,1, 4,7}

So here the elements in the compliment of the given set -2,-1,0,1,4,7

4 0
3 years ago
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