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2 years ago

Which expression simplifies to 41? Choose all that apply!

Mathematics

1 answer:

Arada [10]2 years ago

5 0

Answer:

None of the expressions is simplified to 41.

Step-by-step explanation:

You have to follow the P.E.M.D.A.S., this is:

P:Parentheses first.

E:Exponent next.

M:Multiplication.

D:Division.

(Multiplication and Division is whichever come first, from left to right)

A: Addition.

S: Subtraction.

(Addition and Subtraction is whichever come first, from left to right).

<u><em>We are going to analyze every option:</em></u>

1. 10+ 2.4 - 1

First we have to solve the multiplication:

$10+ 2.4 - 1=10+8-1$

Now we have to solve the addition and subtraction from left to right:

$10+8-1=18-1=17\neq 41$

This is not a correct answer.

2. 10+2.(4-1)

We have to solve the parentheses first:

$10+2.(4-1)=10+2.(3)$

Now the multiplication and then the addition.

$10+2.(3)=10+6=16\neq 41$

This is not a correct answer.

3. (10+23).4-1

We have to solve the parentheses first:

$(10+23).4-1=(33).4-1$

Now the multiplication and then the subtraction:

$(33).4-1=132-1=131\neq 41$

This is not a correct option.

4. $10+(2^2.4)-1$

We have to solve the parentheses first, in the parenthesis first we have to solve: $2^2$ and then do the multiplication.

$10+(2^2.4)-1=10+(4.4)-1=10+16-1$

Now we have to solve the addition and subtraction from left to right:

$10+16-1=26-1=25\neq 41$

This expression is not a correct option.

Then none of the expressions is simplified to 41.

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. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

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Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

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