What is the vertex of the graph of the function below ?<br> y=x^2-4x+3

For every 15 minutes of driving you've drove 600 miles what is the unit rate

ElenaW [278]

Answer:

Step-by-step explanation:

40 miles per minute

4 0

2 years ago

Need help asap with these

IRISSAK [1]

Answer:

1. 6n - 15

2. x + 10

3. -4d + 15

4. true, associative property

5. distributive, commutative

Step-by-step explanation:

1. distribute 3/4 into the parentheses, so multiply 3/4 * 8 and 3/4 * -20

2. distribute the 2 into the parentheses so multiply 2 * 3x and 2 * 5, then combine like terms

3. distribute -3 into the parentheses so -3 * 4D and -3 * -5 then add like terms

4. add like terms on both sides of the equal sign and you get 10x on both sides. this is the associative property because it doesn't matter what's inside or outside the parentheses you get the same answer no matter how you add them up

5. the first part is distributive because you distributed the three into the parentheses and the second part is commutative because you switch them around and you can get the same answer

6 0

3 years ago

I’m stuck on this question can anyone help me?

ch4aika [34]

Answer: REE thats high school work

Step-by-step explanation:

7 0

2 years ago

a large container holds 4 gallons of chocolate milk that has to be poured into bottes. each bottle holds 2 pints.

timama [110]

U can pour it into 16 bottes

5 0

3 years ago

Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

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Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago

What is the vertex of the graph of the function below ? y=x^2-4x+3