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3 years ago

You can know that 236 is a composite number because it is divisible by _____. A. 2 B. 3 C. 7 D. 6

Mathematics

2 answers:

kaheart [24]3 years ago

3 0

A. 2

<h2>Explanation:</h2>

A composite number can be formed if you multiply two smaller positive integers. So the result is also a positive integer. Let's test each option:

Option A.

$\frac{236}{2}=118 \\ \\ \\ So: \\ \\ 236=2\times 118$

236 can be formed by multiplying two smaller positive integers that are 2 and 118.

Option B.

$\frac{236}{3}=78.67 \ It's \ not \ an \ integer$

Option c.

$\frac{236}{7}=33.71 \ It's \ not \ an \ integer$

Option d.

$\frac{236}{7}=39.33 \ It's \ not \ an \ integer$

<em>Only Option A meets the requirements.</em>

<h2>Learn more:</h2>

Inequalities: brainly.com/question/4600160

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saul85 [17]3 years ago

3 0

<u>Answer:</u>

Option(a)

You can know that 236 is a composite number because it is divisible by 2

<u>Explanation:</u>

A composite number is a number which is a positive integer and non prime in nature i.e it is divisible by number other than 1 and itself.

Since the given number 236 is divisible by 2 and gives 118 on dividing by 2

$\frac{236}{2} = 118$

where it an positive integer and it is a composite number.So, option A is the right answer and other option wont result in an integer.

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An element with mass 490 grams decays by 28.6% per minute. How much of the element is remaining after 16 minutes, to the nearest

agasfer [191]

Answer:

2.2 grams

Step-by-step explanation:

We can use the compound decay formula shown below to solve this:

$F=P(1-r)^t$

Where

F is the final amount (what we want to find)

P is the present amount (490 grams)

r is the rate of decay, in decimal (28.6% = 28.6/100 = 0.286)

t is the time in minutes (16)

Substituting, we get:

$F=P(1-r)^t\\F=490(1-0.286)^{16}\\F=490(0.714)^{16}\\F=2.2355$

So, after 16 minutes, 2.2355 grams will be remaining.

<em>Rounded to nearest tenth of a gram (1 decimal place) = </em><em>2.2 grams remaining</em>

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4 years ago

Find all subsets of the set s = {(i, 0), (0, i), (i, i)} that form a basis for r2•

sasho [114]

All subsets of the set s = {(i, 0), (0, i), (i, i)} that form a basis for r2 is

{( 0,1) (-1,1)}

{(1,0)(0,1)}

{(1,0)(-1,1)}

A fixed A is a subset of some other set B if all elements of set A are factors of set B. In other phrases, set A is contained within set B. The subset relationship is denoted as A⊂B.

Subsets are a part of one of the mathematical ideas known assets. a hard and fast is a set of items or factors, grouped within the curly braces, consisting of {a,b,c,d}. If a set A is a set of even numbers and set B includes {2, 4,6}, then B is stated to be a subset of A, denoted by B⊆A and A is the superset of B.

In mathematics, set A is a subset of a fixed B if all factors of A are also elements of B; B is then a superset of A. It's far possible for A and B to be equal; if they may be unequal, then A is a proper subset of B. The relationship of one set being a subset of every other is known as inclusion.

A set which is (is linearly independent andoil generates/ spans the space (like R) is called a baris.

S= (1,0), (0,1), (-1))}

option-1

As, dimension of R²=2, so, Ary set

conists of three vectors cannot be baris for R²

so, option-1 is wrong.

option-2

As, dimention of R2-2.

{(10), (-11)} is a linearly independent

subset of S. so And as, this subset

has 2 vectors so it will as also

span R se, it will be a baris

of IR?

So, option-6 is correct

{(0,1)} is linearly independent subset

option-7

As (LO) has one vector. So, it cannot

Span/generate R so, it also cannot be balis

for R2

of S.

But it cannot span R²

So ut cannot

be a baris of R2

se, option-2 is wrong.

Option-3

AS, {(0,1), (-1,1)} is linearly independent

So, option -7, is wrong.

Answer

[as fox amy KER, (-1,1) K (0,1)] subset of s

And also as dim {(0,1), (-11)}=2

which is same as the dimention of IR2

so {(0,1), (-11)} forms a baris of R2

{(1,0), (0,1), (-11)}

{(1,0)}

So, option-3 is

Connect

option-4

{(1,0), (3, 1)} is standard balis of R2

which is subset of S.

[{(1,0), (0,1)}]

So, option-4 is correct.

option-5

{(-11)}

consists of one vertex,

so, ut cannot span R², so, ut cannot

be a baris of R²

9090

LO {(1,0), (-1,1)}]

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