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2 years ago

Let's find the unknown variables in the following cases. ) (a) If C.P. Rs 150 and profit = Rs 12, find profit percent

Mathematics

2 answers:

vivado [14]2 years ago

6 0

Answer:

solution, cp=Rs150 profit=Rs12 Now, profit percent=profit% of cp or,profit percent=12/100 x Rs150 Therefore,profit percent =18%

Step-by-step explanation:

Romashka [77]2 years ago

3 0

Answer:

Step-by-step explanation:

We know that

Profit% = Profit/CP × 100

=> 12/150 × 100

=> 12/15 × 10

=> 4/5 × 10

=> 8%

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P(x)=(x-2)(x-4)(x-5)

Cerrena [4.2K]

The constants of a polynomial is the term that has no variable attached to it.

<h3>The constant term</h3>

To determine the constant, we simply multiply the constant term in each factor of the polynomial.

So, we have:

<h3 /><h3>Polynomial P(x) = (x-2)(x-4)(x-5)</h3>

$Constant = -2 \times -4 \times -5$

$Constant = -40$

Hence, the constant is -40

<h3>Polynomial P(x) = (x-2)(x-4)(x+5)</h3>

$Constant = -2 \times -4 \times 5$

$Constant = 40$

Hence, the constant is 40

<h3>Polynomial P(x) =1/2(x-2)(x-4)(x+5)</h3>

$Constant = \frac 12 \times -2 \times -4 \times 5$

$Constant = 20$

Hence, the constant is 20

<h3>Polynomial P(x) = 5(x-2)(x-4)(x+5)</h3>

$Constant = 5 \times -2 \times -4 \times 5$

$Constant = 200$

Hence, the constant is 200

$Constant = -5 \times -2 \times -4 \times 5$

$Constant = -200$

Hence, the constant is -200

Read more about polynomials at:

brainly.com/question/2833285

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2 years ago

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erastovalidia [21]

Answer:

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Step-by-step explanation:

Divide (4/3) by 2, then multiply by 5

4 0

3 years ago

Read 2 more answers

A 4" by 6" by 8" rectangular solid is cut by slicing through the midpoint of three adjacent edges. What is the number of inches

Triss [41]

Answer:

22.078

Step-by-step explanation:

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2 years ago

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