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3 years ago

Look at the picture and solve the equations

Mathematics

1 answer:

ch4aika [34]3 years ago

6 0

Answer:

1. a= 7, one solution

2. x=x, infinitely solutions

3. y=3, one solution

4. x=6, one solution

5. no solution

6. k = 2, one solution

Step-by-step explanation:

7a - 6 = 43

7a = 49

one solution

a = 7

3x + 4 = 3x + 4

3x = 3x

x = x

infinitely solutions

4(2y - 4) = 5y + 2

8y - 16 = 5y + 2

3y -16 = 2

3y = 18

one solution

y= 6

multiply 2 on both sides

5x + 6 = x + 30

4x = 24

one solution

x = 6

4e - 19 = 4(e-4)

4e - 19 = 4e -16

-19 ≠ -16

no solution

10k - 2(k+3) = 10

10k -2k - 6 = 10

8k - 6 = 10

8k = 16

one solution

k = 2

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Given function:

$f(x)=3x^3-2x^2-11x+10$

If (x + 2) is a factor then:

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To find a, compare the coefficients of x³:

$\implies a=3$

To find b, substitute the found value of a into the coefficient for x² and compare:

$\implies 2a+b=-2$

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To find c, compare the constants:

$\implies 2c=10$

$\implies c=5$

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$\implies f(x)=(x+2)(3x^2-8x+5)$

Now factor (3x²-8x+5):

$\implies 3x^2-3x-5x+5$

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Therefore the factored function is:

$\implies f(x)=(x+2)(3x-5)(x-1)$

Zero Product Property

$\boxed{\text{If \; $a \cdot b = 0$ \; then either \; $a = 0$ \;or \;$b = 0$ (or both)}.}$

To find the zeros, set each factor equal to zero and solve for x:

$\implies x+2=0 \implies x=-2$

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$\implies x-1=0 \implies x=1$

Therefore, the zeros of the function are:

$x=-2, \quad x=\dfrac{5}{3},\quad x=1$

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