Help with # 4 plz idk

notka56 [123]

The answer is 138
Because the angle is 180 and what I did was 2 x 21 = 42 and I did 180-42 equals to 138

7 0

1 year ago

6. Consider the polynomial p(x) = x^2 + kr^2 +x+6. Find a value of k so that x+1 is a factor of P. Find all the zeros of P.

Nikitich [7]

If x + 1 is a factor of p(x) = x³ + k x² + x + 6, then by the remainder theorem, we have

p (-1) = (-1)³ + k (-1)² + (-1) + 6 = 0 → k = -4

So we have

p(x) = x³ - 4x² + x + 6

Dividing p(x) by x + 1 (using whatever method you prefer) gives

p(x) / (x + 1) = x² - 5x + 6

Synthetic division, for instance, might go like this:

-1 | 1 -4 1 6

... | -1 5 -6

----------------------------

... | 1 -5 6 0

Next, we have

x² - 5x + 6 = (x - 3) (x - 2)

so that, in addition to x = -1, the other two zeros of p(x) are x = 3 and x = 2

6 0

3 years ago

Jessie wants to buy three tickets for a concert. she has earned $135 and each tickey costs $50. she borrows the rest of the mone

dybincka [34]

Answer: In explanation

Step-by-step explanation:

1. 135 - 50x

2. $15

3. $0

I hope this helped!

3 0

1 year ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

Can someone please help me in this question

Pie

Step-by-step explanation:

a) Let y = f(x) = 3x - 2x^2

f(-2) = 3(-2) - 2(-2)^2 = p

= -6 - 8

= -14

= p

f(2.5) = 3(2.5) - 2(2.5)^2 = q

= 7.5 - 12.5

= -5

= q

b) graphing

c) From the graph, you should be able to verify the following:

i) f(0.5) = 3(0.5) - 2(0.5)^2 = 1

ii) 0.5 = 3x - 2x^2 or x = 1.3, 0.2

iii) the maximum occurs at

f(0.75) = 1.125

d) the equation for the line of symmetry is x = 0.75

4 0

3 years ago