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1 year ago

Y= x^2-4x+3 in vertex form

Mathematics

1 answer:

Karolina [17]1 year ago

4 0

Answer:

$y=(x-2)^2-1$

Step-by-step explanation:

<u>Complete the square</u>

<u /> $y=x^2-4x+3\\\\y+1=x^2-4x+3+1\\\\y+1=x^2-4x+4\\\\y+1=(x-2)^2\\\\y=(x-2)^2-1$

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The sum of two numbers is 57. The larger number is 11 more than the smaller number. What are the numbers?

FromTheMoon [43]

Answer:

I'll tell u later

Step-by-step explanation:

Small number =x

Large number =11 plus x

Now from the question,

$x + 11 + x = 57$

Therefore 2x plus 11= 57

2x= 57-11=46

X=46÷2=23

8 0

3 years ago

Logan ate1.438 pounds of grapes his brother Ralph are 1.44 pounds of grapes which brother are more grapes

Lynna [10]

Ralph ate more grapes

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3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!!

podryga [215]

Angle MHL is equal to angle JHN + angle NHK

7 0

2 years ago

HELP!!! if a hospital patient is given 50 milligrams of medicine, which leaves the bloodstream at 10% per hour, how many milligr

kipiarov [429]

Answer:

I also got 27.44 mg.

Step-by-step explanation:

Use the function given to you:

A(t)=le^rt

l: Initial Amount (50)

r: Rate (-.1) (Negative because the medicine is exiting the bloodstream at 10% per hour

t: 6 (the specific time value you are evaluating the function at.

A(6)=50e^-.1(6)=27.44

8 0

2 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

2 years ago