Question

http://prntscr.com/ntaza8

Answer 1

@Narad

Answer 2

Question nº27

\[x^2-9=0\]
Factorising
\[(x+3)(x-3)=0\]
There are 2 solutions
\[x-3=0\]
\[x=3\]
and
\[x+3=0\]
\[x=-3\]
The solutions are
x= {-3,3}
The answer is option B

Answer 3

Question nº28
The function is
\[f(x)=7x^2-28x-60\]
To obtain the vertex, write the functio in the vertex form

Write the function as
\[f(x)=a(x-h)^2+k\]
Therefore,
\[f(x)=7(x^2-4x)-60\]
By completing the squares
\[f(x)=7(x^2-4x+4)-60-28\]
\[f(x)=7(x-2)^2-88\]
The vertex is
\[=(h,k)=(2, -88)\]
The answer is option A

Answer 4

The equation of the height in terms of t is

\[h(t)=-16t^2+64t+80\]
Write the equation in the vertex form
\[h(t)=-16(t^2-4t)+80\]
\[h(t)=-16(t^2-4t+4)+80+64\]
\[h(t)=-16(t-2)^2+144\]
The maximun point is when \[t=2\]
and the height is
\[h(2)=144ft\]
The answer is option A

Answer 5

okay

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Answer 6

The equation of the red line is
\[y=ax+2\]
2 points on the line are
\[(0,2)\] and
\[(2,0)\]
Plugging those values in the equation of the line
\[2=a*0+b\]
and
\[0=2a+b\]
Solving for a and b
\ and \[a=-b/2=-2/2=-1\] The equation of the red line is \[y=-x+2\] The shaded part of the graph is \[y \ge -x+2\] The answer is option A

Answer 7

b=2

Answer 8

http://prntscr.com/ntbc4f

Answer 9

The equation of this graph is
\[y=|x|\]
This is not a quadratic function
The answer is FALSE

Answer 10

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Answer 11

The quadratic function in the vertex form is
\[y=a(x-h)^2+k\]
From the graph, the vertex is
\[=(h,k)= (-2,-3)\]
The k value is
k=-3
This is option C

Answer 12

Okay

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Answer 13

The vertex form of the equation is
\[y=a(x-h)^2+k\]
Here, the vertex is
\[(h,k)=(-4,-6)\]
The equation of the graph is
\[y=a(x+4)^2-6\]
When x=-2, the value of y=-2
Plugging the values in the equation
\[-2=a(-2+4)^2-6=4a-6\]
4a=4
a=1
The equation of the graph is
\[y=(x+4)^2-6\]
The answer is option C

Answer 14

Okay

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Answer 15

Question nº22
The quadratic equation is
\[y^2-7y-18=0\]
Rewriting the equation as
\[y^2-7y-2y+2y-18=0\]
\[y^2-9y+2y-18=0\]
\[y(y-9)+2(y-9)=0\]
Factorising
\[(y+2)(y-9)=0\]
The solutions are
y+2=0
y=-2
and
y-9=0
y=9
The solutions are
S={-2,9}
The answer is option A

Answer 16

Question nº23
\[x^2-56=10x\]
Rewriting the equation as
\[x^2-10x-56=0\]
\[x^2-10x-4x+4x-56=0\]
\[x^2-14x+4x-56=0\]
\[x(x-14)+14(x-14)=0\]
Factoring
\[(x+4)(x-14)=0\]
The solutions are

x+4=0
and
x-14=0

x=-4 and a=14

The solutions are

S={-4,14}
The answer is option D

Answer 17

Question nº24
Are you familiar with the discriminant?

The equation is
\[2x^2-3x-5=0\]
Comparing this to the equation
\[ax^2+bx+c=0\]

a=2
b=-3
c=-5

The discriminant is
\[\Delta =b^2-4ac =(-3)^2-4*(2)*(-5)=9+40=49\]
As the discriminant is
\[\Delta >0\]
There are 2 solutions
The answer is option C

Answer 18

Okay

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Answer 19

Question nº25

The equation is
\[x^2-5x+9=0\]
a=1
b=-5
c=9

The discriminant is
\[\Delta = b^2-4ac=(-5)^2-4*1*9=25-36=-9\]
as
\[\Delta < 0\]
There are no real solutions

The solutions are
\[x=(-b \pm \sqrt{\Delta})/(2a) = (5\pm \sqrt{-9})/(2*1)\]
\[x=5/2\pm 3/2i\]

I think there is a problem with the solutions proposed.

Answer 20

Question nº26
The equation is

\[x^2-6x+4=0\]

a=1
b=-6
c=4

The discriminant is
\[\Delta= b^2-4ac=(-6)^2-4*4=36-16=20\]
The solutions are
\[x=(-b \pm \sqrt{\Delta})/(2a)=(6\pm \sqrt{20})/(2) =(6\pm2\sqrt{5})/2=3\pm \sqrt{5}\]
This is TRUE

Answer 21

Question nº27
\[x^2-9=0\]
Factorising
\[(x-3)(x+3)=0\]
The solutions are
x+3=0
and
x-3=0

x=-3 and x=3

The solutions are

S={-3, 3}
The answer is option B

Answer 22

Okay

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ignore my answers

Answer 23

We have already replied to those questions

Answer 24

No I accidentally put those answers but we did not do them yet

Answer 25

oh wait sorry

Answer 26

I got it my bad

Answer 27

http://prntscr.com/ntc10h

Answer 28

Question nº 28 was the second question
Question nº29 was the third question

Answer 29

Yes I got that

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Answer 30

This is a quadratic function

The domain is all real values
\[x \in \mathbb{R} \]

The range is
\[y \in [-\infty, 4)\]

Therefore
\[y \le 4\]
The answer is option C

Answer 31

http://prntscr.com/ntc3id

Answer 32

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Narad
Question nº25

@narad attention please 25-36 not equal -9

The equation is
\[x^2-5x+9=0\]
a=1
b=-5
c=9

The discriminant is
\[\Delta = b^2-4ac=(-5)^2-4*1*9=25-36=-9\]
as
\[\Delta < 0\]
There are no real solutions

The solutions are
\[x=(-b \pm \sqrt{\Delta})/(2a) = (5\pm \sqrt{-9})/(2*1)\]
\[x=5/2\pm 3/2i\]

I think there is a problem with the solutions proposed.
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 33

The quadratic equation in vertex form is
\[y=a(x-h)^2+k\]

and the vertex is (12, -8)
\[y=a(x-h)^2+k=a(x-12)^2-8\]
\[y=a(x^2-24x+144)-8=0\]
The zeros are x=10 and x=14

Therefore,
\[0=a(10-12)^2-8=4a-8\]
a=2

\[0=a(14-12)^2-8=4a-8\]
a=2

The quadratic equation is
\[y=2(x-12)^2-8\]

\[y=2(x^2-24x+144)-8=2x^2-48x+280\]
The answer is option A

Answer 34

Correction question nº25
The discriminant is
\[\Delta=25-36=-11\]
The solutions are
\[x=(5\pm \sqrt{11}i)/2=5/2 \pm \sqrt{11}/2i\]
The answr is option B

Answer 35

Question nº32
The vertex form of a quadratic equation is
\[y=a(x-h)^2+k\]

The vertex is = (17,-2)

Therefore,

\[y=a(x-17)^2-2\]

The zeros of the equation are
x=16
and
x=18
Plugging those values
\[0=a(16-17)^2-2=a-2\]
a=2
\[0=a(18-17)^2-2=a-2\]
a=2
The equation is
\[y=2(x-17)^2-2= 2(x^2-34x+289)-2=2x^2-68x+576\]
The answer is option B

Answer 36

Okay

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Answer 37

Let the width of the deck = x yd

The area of the pool is = 15*20=300 yd^2

The new area of the pool is = 600 yd^2

The new area is

\[A= (15+2x)(20+2x)= 300+70x+4x^2=600\]
\[4x^2+70x-300=0\]
\[2x^2+35x-150=0\]
a=2
b=35
c=-150
The solutions to this quadratic equation are

\[x=(-35\pm \sqrt{35^2-4*2*(-150)})/(2*2)=(-35\pm \sqrt{2425})/4=(-35\pm49.2)/4\]
We keep the positive value
\[x=(-35+49.2)/4=3.56 \approx 3.6\]
The answer is option A

Answer 38

http://prntscr.com/ntcc2t

Answer 39

The solution is
\[16^{-3/4} = (1/16)^{3/4} = ((1/16)^{1/4})^{3}=(1/2)^{3}=1/8\]
The answer is false since the expression is not =1/4 but 1/8

Answer 40

http://prntscr.com/ntk6ph

Answer 41

@Narad