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3 years ago

Which point is the vertex of the quadratic function ()=2−6+8?

Mathematics

1 answer:

xenn [34]3 years ago

8 0

Answer:

Step-by-step explanation:

1.Get the equation in the form y = ax2 + bx + c.

2.Calculate -b / 2a. This is the x-coordinate of the vertex.

3.To find the y-coordinate of the vertex, simply plug the value of -b / 2a into the equation for x and solve for y.

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Latrell took one of his friends out to lunch. The lunches cost $60 and he paid 5% sales tax. If Latrell left a 15% tip on the $6

VladimirAG [237]

Answer:

$72

Step-by-step explanation:

Cost of the lunch = $60

Sales tax = 5% of $60

= 5/100 × 60

= 0.05 × 60

= $3

Tip = 15% of $60

= 15/100 × 60

= 0.15 × 60

= $9

how much in total did he pay?

Total amount paid = cost of the lunch + sales tax + tip

= $60 + $3 + $9

= $72

Latrell paid $72 in total for the lunch

4 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago

Find the image of (2,-7) after a reflection over the x-axis

asambeis [7]

Answer: (2, 7)

Step-by-step explanation:

To reflect something over the x-axis, simply invert the y value. Simply turn -7 into 7.

Hope it helps <3 :)

3 0

3 years ago

The probability of Andy passing his driving test

ivann1987 [24]

Answer:

70%

Step-by-step explanation:

100% - 30% = 70%

3 0

3 years ago

Read 2 more answers

How many millimetres are there in 5.5 litres?

kykrilka [37]

$l-liter\\mm-milimeter\\dm-decimeter\\--------------------\\deci\ (d)=0.1\ of\ some\ value\\milli\ (m)=0.001\ of\ some\ value\\-----------------\\therefore\ 1dm=100mm\\-----------------\\1l=1dm^3\\1dm^3=(100mm)^3=100^3mm^3=1,000,000mm^3\\---------------------\\Your\ a\ question:\\\\5.5l=5.5dm^3=5.5\cdot1,000,000mm^3=5,500,000mm^3$

$Scientific\ notation:\\\\5,\underbrace{500,000}_{\leftarrow6}mm^3=5.5\cdot10^6mm^3$

4 0

3 years ago

Read 2 more answers