Putting money into more than one kind of investment at a time is called

PLS HELP!

wolverine [178]

Given:

Alexis has a rectangular backyard that is 50 yards by 55 yards.

She wants to build a fence that stretches diagonally from one corner to the opposite corner.

To find:

The length of the fencing she needs.

Solution:

We have,

Length = 50 yards

Width = 55 yards

We know that, the diagonal of a rectangle is

$Diagonal=\sqrt{length^2+width^2}$

$Diagonal=\sqrt{(50)^2+(55)^2}$

$Diagonal=\sqrt{2500+3025}$

$Diagonal=\sqrt{5525
}$

On further simplification, we get

$Diagonal=74.3303437$

$Diagonal\approx 74.3$

Therefore, the length of the required fencing is 74.3 yards.

8 0

2 years ago

Which of the following circles have their centers on the Y axis

mixer [17]

Answer:

Hi, please provide a photo or options to pick from. I will answer then.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Greatest president ~ A Gallup Poll of American adults in February 2009 asked the following question. "Which of the following pre

Anastasy [175]

Answer:

The sample proportion is 0.19815

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound. The sample proportion is the halfway point between these two bounds, that is, the sum of the bounds divided by 2.

In this problem, we have that:

Lower bound: 0.1759

Upper bound: 0.2204

Sample proportion:

(0.1759 + 0.2204)/2 = 0.19815

The sample proportion is 0.19815

6 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

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

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

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

ABC IS a right angled triangle at B and D is a point on BC. If AD=18cm , BD=9cm and CD = 4 cm, find AC

leva [86]

$AB = \sqrt{ AD^{2} - BD^{2} } = \sqrt{243} = 9 \sqrt{3} 
$
$AC = \sqrt{ AB^{2} + BC^{2} } = \sqrt{412} = 2 \sqrt{103}$

4 0

3 years ago