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VashaNatasha [74]
2 years ago
14

System A

Mathematics
1 answer:
Sonbull [250]2 years ago
4 0

For the given systems we have:

1) Swap the right sides of both equations in system A.

2) No, the systems are not equivalent systems.

<h3>How can we get System B from System A?</h3>

The two systems are:

A:

-10x - 4y = 0

-2x + 7y = 8

B:

-10x - 4y = 8

-2x + 7y = 0

Notice that the only different thing between the systems is that the right sides are inverted. So, if we invert the right side of A, we will get system B.

2) Are the systems equivalent?

No, the systems are not equivalent because to go from system A to system B we only modified the right side of system A, the systems would be equivalent only if modifying both sides of A gave system B, but this is not the case.

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

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Ben had $100 in his savings account. He adds $5.00 a week to his savings as shown in the graph below. According to this graph, h
bearhunter [10]

Answer:

Option (C)

Step-by-step explanation:

Let the relation between amount of money (m) and Time (t) is represented by the equation,

m = ct + b

Where c = Investment per week

b = Ben already had the amount before investment

From the question,

c = $5 per week

b = $100

Therefore, equation will be,

m = 5t + 100

If duration of the investment 't' = 8 weeks

Amount in the account after 8 weeks,

m = 5(8) + 100

m = $140

Option (C) is the correct option.

3 0
3 years ago
What’s the answer of a multiplying polynomials (3x-1) to the second power and (x+6) to the second power
Vladimir [108]

9x^4+102x^3+253x^2-204x+36 is the result of multiplying (3x-1) to the second power and (x+6) to the second power

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

Given that we have to find the result of multiplying polynomials (3x-1) to the second power and (x+6) to the second power

"Second power" means the term is raised to power of 2

Therefore,

We have to multiply (3x-1)^2 \text{ and }(x+6)^2

\rightarrow (3x-1)^2 \times (x+6)^2

We can use the algebraic identity to expand the above expression

(a+b)^2 = a^2+2ab+b^2\\\\(a-b)^2=a^2-2ab+b^2

Applying these in above expression, we get

\rightarrow ((3x)^2-2(3x)(1)+1^2) \times (x^2+2(x)(6)+6^2)\\\\\rightarrow (9x^2-6x+1) \times (x^2+12x+36)

Multiply each term in first bracket with each term in second bracket

\rightarrow 9x^2(x^2)+(9x^2)(12x)+(9x^2)(36)-6x(x^2)-6x(12x)-6x(36) + x^2+12x+36

Simplify the above expression

\rightarrow 9x^4+108x^3+324x^2-6x^3-72x^2-216x+x^2+12x+36

Combine the like terms

\rightarrow 9x^4+102x^3+253x^2-204x+36

Thus the above expression is the result of multiplying (3x-1) to the second power and (x+6) to the second power

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3 years ago
HELP PLZ IDK THE ANSWER<br> 54+6
viva [34]

Answer:

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Step-by-step explanation:

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(a^3+14a^2+33a-20)\(a+4)
Oksi-84 [34.3K]
Perhaps you meant   <span>(a^3+14a^2+33a-20)   /   (a+4), for division by (a+4).  

Do you know synthetic division?  If so, that'd be a great way to accomplish this division.  Assume that (a+4) is a factor of </span>a^3+14a^2+33a-20; then assume that -4 is the corresponding root of a^3+14a^2+33a-20.

Perform synth. div.  If there is no remainder, then you'll know that (a+4) is a factor and will also have the quoitient.

-4  /   1   14   33   -20
     ___    -4_-40    28___________
          1   10   -7       8

Here the remainder is not zero; it's 8.  However, we now know that the quotient is 1a^2 + 10a - 7 with a remainder of 8.
8 0
3 years ago
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