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3 years ago

Are these expressions the same c = 1cc + 1/4c why or why not

1 answer:

6 0

Yes 1c = c.

Technically, the "1" in front of the "c" is called an "implied 1" and, so it's simply written as c.

Here's, an example that should make this clear:
Say c = 4
Then (1)(c) = (1)(4) = 4 = c

1c + 1/4 c

What is a whole plus 1/4? 1 1/4 or 5/4
Same thing here:
c + 1/4 c or 1c + 1/4 c = 1 1/4c or 5/4c

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100 POINTS HELP ASAP Linear Combination

notka56 [123]

Answer:

1) $x=\dfrac12$

2) $n = -3 \ \ \textsf{and} \ \ m = -4$

3) see below

4) A: 0 = 1

Step-by-step explanation:

<u>Question 1</u>

$15-0.5(4x-2)+4x=17$

$\implies 15-2x+1+4x=17$

$\implies 2x+16=17$

$\implies 2x=1$

$\implies x=\dfrac12$

<u>Question 2</u>

$\textsf{rearrange} \ n=m+1 :$

$\implies -m=-n+1$

$\textsf{add equations} \ -m=-n+1 \ \textsf{and} \ m=2n+2:\\$

$\implies0=n+3$

$\implies n=-3$

$\textsf{substitute} \ \ n=-3 \ \ \textsf{into} \ \ m=n-1:$

$\implies m=-3-1$

$\implies m=-4$

<u>Question 3</u>

subtract the second equation from the first

divide both sides by -4

substitute found value for y into first equation

solve for x

<u>Question 4</u>

$3j=k$

$k=3j+1$

$\textsf{rearrange} \ 3j=k :$

$\implies -k=-3j$

$\textsf{add equations}\ -k=-3j \ \ \textsf{and}\ \ k=3j+1:$

$\implies 0=1$

Solution = A

6 0

3 years ago

How_many_meatballs_are_required_for_a_party_of_25_adults_and_6_children

Tanya [424]

Meatballs required 13

7 0

3 years ago

Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?

-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>

In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

U = {x | x is a real number}
A = {x | x is an odd integer}
R = {x | x = 3, 7, 11, 27}

So,

A = 1, 3, 5, 7, 9, 11... are the elements of set A.
R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(D) no, because each element in set R is not represented in set A

8 0

2 years ago

Brianliest help please

Mamont248 [21]

Answer:

answer is option C

3+7=10

7-3=4

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME! I BEG YOU! ILL EVEN GIVE BRAINLIEST!! HELP ASAP!!!

andreev551 [17]

D because u have to look the fraction

6 0

3 years ago