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3 years ago

A concrete patio is 5 2/3 feet wide. It has an area of 36 5/6 square feet. What is the length of the concrete patio

Mathematics

1 answer:

Aneli [31]3 years ago

4 0

The length of the concrete patio is 6 1/2.
I got this answer by dividing 36 5/6 and 5 2/3. Therefore, answer is 6 1/2

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Hey guys umm help me

gavmur [86]

The answer is c

1/6 divided by 1/2

i had this question on my own

3 0

3 years ago

If there are about 1 2/5 kilometers in 1 mile,about how many kilometers are in 4 miles?

-BARSIC- [3]

This problem is an example of ratio and proportion. A ratio is a comparison between two different things. You are given the equivalent distance of a 1 2/5 kilometers to 1 mile. Also you are given 4 miles. You are required to find the distance in kilometers of 4 miles. The solution of this problem is,

1 2/5 kilometers /1 mile = distance/ 4 miles
distance = (4 miles) (1 2/5 kilometers /1 mile)
distance = 28/5 or 5.6 kilometers
There are 5.6 kilometers in 4 miles.

7 0

3 years ago

Pls help i will provide picture

VashaNatasha [74]

Answer:

A. 4

Step-by-step explanation:

First, we can try the first option. 2/4 is equivalent to 6/12 when you multiply the fraction by 3 to get a common denominator. Since 6/12 is larger than 4/12, and the question has 1 answer, 4 is the only value that satisfices the statement and is the answer.

6 0

2 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

3 years ago

Is the following relation a function? Two circles are shown, one labeled x and the other labeled y. The x circle contains the nu

Allushta [10]

Answer:

i believe that it is a function.

Step-by-step explanation:

you do not have any of the exact same numbers with arrows ot one another.

5 0

3 years ago