2 years ago

The incorrect work of a student to solve an equation 2(y + 4) = 4y is shown below:

Mathematics

1 answer:

yulyashka [42]2 years ago

8 0

Answer:

3; y=4

Step-by-step explanation:

2 should be distributed as 2y + 8; y = 4

2(y +4) = 4y

2y + 8 = 4y

8 = 2y

y = 4

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If f(x)=x^2find f(2)-f(0)/2

alukav5142 [94]

Answer:

Step-by-step explanation:

On the picture above.

8 0

3 years ago

Which correctly describes the roots of the following cubic equation x^3-5x^2+3x+9=0? A. Three real roots, each with a different

77julia77 [94]

Answer:

The correct option is C.

Step-by-step explanation:

The given cubic equation is

$x^3-5x^2+3x+9=0$

According to the rational root theorem 1 and -1 are possible rational roots of all polynomial.

At x=-1, the value of function 0. Therefore (x+1) is the factor of polynomial and -1 is a real root.

Use synthetic division to find the remaining polynomial.

$(x+1)(x^2-6x+9)=0$

$(x+1)(x^2-2(3)x+3^2)=0$

Using $(a-b)^2=a^2-2ab+b^2$

$(x+1)(x-3)^2)=0$

USe zero product property and equate each factor equal to 0.

$x=-1,x=3,x=3$

Therefore the equation have three real roots out of which the value of two roots are same.

Option C is correct.

7 0

3 years ago

Read 2 more answers

05) A polynomial function f of degree(6) whose coefficients are real numbers has the zeros i, 4-i,2+i . Find the remaining zeros

arlik [135]

The degree of the polynomial function f is the number of zeros function f has.

The remaining zeros of the polynomial function are -i, 4 + i and 2 - i

<h3>How to determine the remaining zeros</h3>

The degrees of the polynomial is given as;

Degree = 6

The zeros are given as:

i, 4-i,2+i

The above numbers are complex numbers.

This means that, their conjugates are also zeros of the polynomial

Their conjugates are -i, 4 + i and 2 - i

Hence, the remaining zeros of the polynomial function are -i, 4 + i and 2 - i