Which number produces an irrational number when added to 1/2?

On the world map,the scale is 1/5= 636 miles.The distance is 2 and 1/8 inches.How many miles is it?

crimeas [40]

The answer for this problem is that there is 6757.5 miles for 2 and 1/8 inches

6 0

3 years ago

Donald's Earnings, x (dollars) Fallons Earnings, y 38 45 43 50 48 55 53 60 58 65 Which equation best represents Fallon's earning

kherson [118]

Given:

The table of values for Fallon's earnings in terms of Donald's earnings.

To find:

The equation that best represents Fallon's earnings in terms of Donald's earnings.

Solution:

In the given table the x-values are increasing by 5 units and y values increasing by 5 units. It means the rate of change of y with respect to x is constant and the table represents a linear function.

If the graph of a linear function passes through two points, then the equation of linear function is

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

Consider any two points from the table. Let the two points are (38,45) and (43, 50). Then, the equation is

$y-45=\dfrac{50-45}{43-38}(x-38)$

$y-45=\dfrac{5}{5}(x-38)$

$y-45=1(x-38)$

$y-45=x-38$

Adding 45 on both sides, we get

$y=x-38+45$

$y=x+7$

Therefore, the correct option is A.

8 0

2 years ago

Area of a pentagon with a radius of 8m

san4es73 [151]

Area = [radius^2 * 5 * sine (360/5)] / 2
area = [64 * 5 * 0.95106] / 2
area = 152.1696

4 0

3 years ago

How tall is the green character?

vodomira [7]

My guess......... um 5'4

5 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

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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