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Zarrin [17]
3 years ago
8

Kiran says that a solution to the equation x+4=20 must also be a solution to the equation 5(x+4)=100 Write a convincing explanat

ion as to why this is true
Mathematics
1 answer:
Helen [10]3 years ago
8 0

Answer:

x+4=20 is a solution to 5(x+4)=100.

Step-by-step explanation:

Following PEMDAS, divide 5 from both sides of the equation 5(x+4)=100, and you have x+4=20 remaining.

This means that both equations have the same solution.

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Read 2 more answers
Q(a+1) - 2q(a) if q(x) = x²+3x+4​
siniylev [52]

q(x) =  {x}^{2} + 3x + 4

_________________________________

Step(1)

To find q(a) we just need to put a instead of x in q(x) function.

Let's do it...

q(a) =  {a}^{2} + 3a + 4

Multiply sides by -2 :

- 2q(a) =  - 2( {a}^{2} + 3a + 4)

- 2q(a) =  - 2 {a}^{2} - 6a - 8

_________________________________

Step (2)

To find q(a+1) we just need to put a+1 instead of x in q(x) function.

Let's do it...

q(a + 1) =  ({a + 1})^{2} + 3(a + 1) + 4 \\

q(a + 1) =  {a}^{2} + 2a + 1 + 3a + 3 + 4 \\

q(a + 1) =  {a}^{2} + 5a + 8

_________________________________

Step (3)

q(a + 1) - 2q(a) =

{a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 =  \\

-  {a}^{2} - a  =  - a(a + 1)

And we're done.

Thanks for watching buddy good luck.

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8 0
3 years ago
The graph below shows two polynomial functions, f(x) and g(x): Graph of f of x equals x squared minus 2 x plus 1. Graph of g of
IceJOKER [234]

9514 1404 393

Answer:

  (c)  f(x) is an even degree polynomial with a positive leading coefficient.

Step-by-step explanation:

The leading terms of the two functions are ...

  f(x): x² (even degree, positive coefficient: 1)

  g(x): x³ (odd degree, positive coefficient: 1)

Then it is true that ...

  f(x) is an even degree polynomial with a positive leading coefficient

8 0
3 years ago
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