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Dvinal [7]
2 years ago
11

Please help i dont get it

Mathematics
1 answer:
Vaselesa [24]2 years ago
5 0

You are being asked to compare various expressions to the given one, and to determine which are equivalent and which are not. You are asked to simplify the given expression—collect terms.

The given expression ...

... 4y -8x² -5 +14x² +y -1

can be simplified by identifying like terms and adding their coefficients.

... y(4 +1) +x²(-8 +14) +(-5 -1)

... = 5y +6x² -6 . . . . . simplified form

Any expression that has a different y-term, a different x² term, or a different constant term is <em>not equivalent</em>.

Once you have found this simplified expression, you can drag it to the appropriate box. Looking at the top three expressions on the left, you see immediately that they have different y-terms, so all those go to the "not equivalent" box. The expression on the bottom row has a different x² term, so it, too, is "not equivalent". (The sign is negative instead of positive. Details matter.)

The remaining expression, the one on the far right, has the appropriate y-term and constant term. The x² terms have not been combined, so it is equivalent, but not fully simplified.

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Kamal and Anand each Lent the same sum of money for 2 years at 5 percent at simple interest and compound interest respectively.
Ivenika [448]

Answer:

  • amount lent: ₹6000
  • interest received: Kamal, ₹600; Anand, ₹615.

Step-by-step explanation:

For principal P invested at simple interest rate r, the returned value in t years is ...

  A = P(1 +rt)

If K is Kamal's returned value, the given numbers tell us ...

  K = P(1 +0.05·2) = 1.1P

__

For principal P invested at compound interest rate r, with interest compounded annually for t years, the returned value is ...

  A = P(1 +r)^t

If A is Anand's returned value, the given numbers tell us ...

  A = P(1.05)² = 1.1025P

This latter amount is RS.15 more than the former one, so we have ...

  1.1025P = 1.1P +15

  0.0025P = 15 . . . . . . . . subtract 1.1P

  P = 6000 . . . . . . . . . . . divide by 0.0025 . . . .  the amount lent

Kamal received 1.1P -P = 0.1P = 600 on the investment.

Each lent ₹6000. Kamal received ₹600 in interest; Anand received ₹615 in interest.

4 0
3 years ago
To sell 300 notebooks and 120 pens quickly, a store decided to create two types of gift sets. one set is to include 2 notebooks
VikaD [51]

The store must sell 60 of each set each to obtain the biggest possible income

<h3>How to determine the number of each gift set?</h3>

The given parameters can be represented using the following table:

                        <u>Set 1        Set 2     Available</u>

Notebook         2                3         300

Pen                   1                  1          120

Selling price     8               11.5

Using the above data values, we have:

Objective function: Max P = 8x + 11.5y

Subject to:

2x + 3y ≤ 300

x + y ≤ 120

Next, we plot the graph of the above inequalities (see attachment)

From the attached graph, we have:

(x,y) = (60,60)

Hence, the store must sell 60 of each set each to obtain the biggest possible income


Read more about maximizing functions at:

brainly.com/question/16826001

#SPJ1

5 0
1 year ago
Plz help dicriminant of 3x^2-4x+2
Zinaida [17]

Answer: it's -8

Step-by-step explanation: use the values of a, b, and c to find the discriminant. give me brainliestt

7 0
3 years ago
Read 2 more answers
In which quadrant or on which axis would you find the point <br> ( 2, 7 )
hram777 [196]

Answer:

The answer is Quadrant 1

7 0
3 years ago
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Prove that x-1 is a factor of x^n-1 for any positive integer n.
PolarNik [594]

Answer:    

x-1 is a factor of x^n - 1

Step-by-step explanation:

x-1 is a factor of x^n - 1

We will prove this with the help of principal of mathematical induction.

For n = 1, x-1 is a factor x-1, which is true.

Let the given statement be true for n = k that is x-1 is a factor of x^k - 1.

Thus, x^k - 1 can be written equal to  y(x-1), where y is an integer.

Now, we will prove that the given statement is true for n = k+1

x^{k+1} - 1\\=(x-1)x^k + x^k - 1\\=(x-1)x^k + y(x-1)\\(x-1)(x^k + y)

Thus, x^k - 1 is divisible by x-1.

Hence, by principle of mathematical induction, the given statement is true for all natural numbers,n.

3 0
3 years ago
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