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3 weeks ago@Narad

3 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoThe 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 weeks agoThe 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**

b=2

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

3 weeks agoThe 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 weeks agoThe 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 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoQuestion 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 weeks agoWe have already replied to those questions

3 weeks agoNo I accidentally put those answers but we did not do them yet

3 weeks agooh wait sorry

3 weeks agoI got it my bad

3 weeks agoQuestion nº 28 was the second question Question nº29 was the third question

3 weeks agoThis 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 weeks ago\(\color{#0cbb34}{\text{Originally Posted by}}\) @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 weeks agoThe 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 weeks agoCorrection 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 weeks agoQuestion 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 weeks agoLet 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 weeks agoThe 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 weeks ago@Narad

3 weeks ago