Alvin cuts 3/4 of peice of cheese. He gives 1/8 of it to Matt. How much of the cheese does Alvin give to Matt?

Give the coordinates of a point on the line whose equation in point-slope form is.

pantera1 [17]

Answer:

(4, -2)

Step-by-step explanation:

The point-slope form of the equation for a line is ...

y -k = m(x -h)

for a line with slope m through point (h, k).

Comparing this to the equation you're given, you can see that the point that was used is (h, k) = (4, -2).

_____

You can find other points on the line, but this one is the easiest to find, since it can be read directly from the equation.

7 0

3 years ago

Is the ratio 2.5:3.5 proportional to the ratio 12:5

Helga [31]

No. 2.5:3.5 would be either proportional to 12:16.8 or 3.571428( which is 25/7);5

8 0

3 years ago

. Which function's graph has a vertex at (-1, -3) and contains the point (2, 15)?<br> HELPPP PLEASEE

AlexFokin [52]

Answer:

2(x+1)²-3

Step-by-step explanation:

Using the vertex formula we can write the follwoing equation

y=a(x+1)²-3

We need to solve for a

we know that when x= 2 y = 15

So we can write

15=a(2+1)²-3

15=9a-3

18=9a

a=2

Which means our final equation is

2(x+1)²-3

in standard form (just in case)

2x²+4x-1

5 0

3 years ago

True or false?

Firdavs [7]

Answer:

b

Step-by-step explanation:

brainliest pls

8 0

3 years ago

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

============================================================

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)

-------------------------------------------

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).

-------------------------------------------

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

3 years ago