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BartSMP [9]
2 years ago
12

when solving the equation 5x-3=5, MaReeco rewrote the equation as 5x =2.Did MaReeco perform an acceptable move to this equation?

why or why not?​
Mathematics
1 answer:
aliina [53]2 years ago
6 0

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

5x-3=5

MaReeco's rewrote equation is given by

5x = 2

As MaReeco subtracted 3 to the right side of the equation, which imbalanced the equation. Hence, it is not correct.

Correction:

For the next step, MaReeco should have added 3 to both sides

i.e.

5x-3+3=5+3

5x=8

Therefore, the acceptable move would have been:

  • 5x=8
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Two equations are given below:
ycow [4]
Hello,

a-3b=4 (1)
a-b=-2 (2)

(1)-(2)==> -2b=6
=>b=-3 and a=-2+(-3)=-5

Answer D


Proof
-5+3=-2
-5-3(-3)=4



8 0
3 years ago
Use differentiation rules to find the values of a and b that make the function f(x) = ( x 2 if x ≤ 2, ax3 + bx if x > 2 diffe
Maurinko [17]

Answer:

The values of a and b are  a=\frac{1}{4} and b = 1

Step-by-step explanation:

* <em>Lets explain how to solve the equation</em>

  f(x) =  {x²                x  ≤ 2

            {ax³ + bx     x > 2  

* We need to find the values of a , b that make the function

 differentiable at x = 2

- <em>At first for f(x) to be continuous at x = 2, substitute x by two in the</em>

<em>  the two expressions and equate them</em>

∵ f(x) = x² at x ≤ 2 and f(x) = ax³ + bx at x > 2

∴ f(2) = (2)² = 4 ⇒ (1)

∴ f(2) = a(2)³ + b(2)

∴ f(2) = 8a + 2b ⇒ (2)

- Equate (1) and (2)

∴ 8a + 2b = 4 ⇒ (3)

* <em>For f(x) to be differentiable when x = 2, the function must be </em>

<em>  continuous when x = 2 and the one-sided derivatives must be </em>

<em>  equal when x = 2</em>

# <u>Remember</u>: If f(x)=ax^{b} , then f'(x)=abx^{b-1}

 If f(x)=ax , then f'(x)=a

 If f(x)=a , then f'(x)=0

∵ f(x) = x²

∴ f'(x) = 2x

- Substitute x by 2

∴ f'(2) = 2(2) = 4

∴ f'(2) = 4 ⇒ (4)

∵ f(x) = ax³ + bx

∴ f'(x) = 3ax² + b

- substitute x by 2

∴ f'(2) = 3a(2)² + b

∴ f'(2) = 12a + b ⇒ (5)

- Equate (4) and (5)

∴ 12a + b = 4 ⇒ (6)

* Now we have system of equations

 8a + 2b = 4 ⇒ (3)

 12a + b = 4 ⇒ (6)

- Multiply equation (6) by -2 to eliminate b

∴ -24 a - 2b = -8 ⇒ (7)

- Add equations (3) and (7)

∴ -16a = -4

- Divide both sides by -16

∴ a = \frac{1}{4}

- substitute the value of a in equation (6)

∴ 12(\frac{1}{4})+b=4

∴ 3 + b = 4

- Subtract 3 from both sides

∴ b = 1

* The values of a and b are  a=\frac{1}{4} and b = 1

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