Mathematics
Nicole:

http://prntscr.com/ntaza8

3 months ago
Nicole:

3 months ago

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

3 months ago

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

3 months ago

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

3 months ago
Nicole: 3 months ago

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

3 months ago

b=2

3 months ago
Nicole:

http://prntscr.com/ntbc4f

3 months ago

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

3 months ago
Nicole:

http://prntscr.com/ntbcjb

3 months ago

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

3 months ago
Nicole: 3 months ago

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

3 months ago
Nicole: 3 months ago

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

3 months ago

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

3 months ago

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

3 months ago
Nicole: 3 months ago

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.

3 months ago

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

3 months ago

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

3 months ago
Nicole:

3 months ago

We have already replied to those questions

3 months ago
Nicole:

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

3 months ago
Nicole:

oh wait sorry

3 months ago
Nicole:

3 months ago
Nicole:

http://prntscr.com/ntc10h

3 months ago

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

3 months ago
Nicole:

Yes I got that http://prntscr.com/ntc10h

3 months ago

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

3 months ago
Nicole:

http://prntscr.com/ntc3id

3 months ago
jhonyy9:

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @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. $$\color{#0cbb34}{\text{End of Quote}}$$

3 months ago

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

3 months ago

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

3 months ago

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

3 months ago
Nicole: 3 months ago

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

3 months ago
Nicole:

http://prntscr.com/ntcc2t

3 months ago

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

3 months ago
Nicole:

http://prntscr.com/ntk6ph

3 months ago
Nicole: