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Setler79 [48]
2 years ago
12

Division w/ remainder help

Mathematics
2 answers:
andrezito [222]2 years ago
8 0

Answer:

If you round up you can buy 5 pizzas and if you don't you can only buy 4 pizzas.  7 and a half bins can hold books. But if you round up you can say that 8 bins are needed to hold all 150 books.

Allisa [31]2 years ago
5 0

Answer:

So for the first one, circle Ignored it becaue you cna't buy any more.

For number 2, Quotient is 8 & R is 10. Circle Rounded the quotient up because you need another bin to hold it.

Step-by-step explanation:

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William works at a nearby electronics store. He makes a commission of 6% percent on everything he sells. If he sells a computer
Naddika [18.5K]

Commission earned by William is $ 45.84

<h3><u>Solution:</u></h3>

Given that,

William makes a commission of 6% percent on everything he sells

He sells a computer for $764.00

<em><u>To find: Commission amount of William</u></em>

Given that he makes a commission of 6 % on everything he sells. So he has received a commission of 6 % of $ 764.00

Commission amount of William = 6 % of $ 764.00

a % of b can be written in fraction as \frac{a}{100} \times b

6 \% \text { of } 764=\frac{6}{100} \times 764=0.06 \times 764=45.84

Thus commission earned by William is $ 45.84

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3 years ago
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Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that A=J^{-1}AJ. Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
4 years ago
How to find cos2theta
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3 years ago
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two hundred tickets for the school play were sold. tickets cost $2 for students and $3 for adults. the total amount collected wa
Gre4nikov [31]
A system of equations is good for a problem like this.
Let x be the number of student tickets sold
Let y be the number of adult tickets sold
x + y = 200
2x + 3y = 490
x = 200 - y
2(200 - y) + 3y = 490
400 - 2y + 3y = 490
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The number of adult tickets sold was 90.
x + 90 = 200 --> x = 110
2x + 3(90) = 490 --> 2x + 270 = 490 --> 2x = 220 --> x = 110
The number student tickets sold was 110.
4 0
3 years ago
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