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qaws [65]
3 years ago
11

Carol is painting a four square game court on a school playground. The four small squares are all the same size. If the total ar

ea of the four small squares is 256 ft2, what is the side length of each small square?
Mathematics
2 answers:
Sliva [168]3 years ago
7 0
Given that the total area of the four small squares is 256 ft2, the first thing we are going to do is to divide 256 by 4 and we get 64. So the area of each small square is 64. Now, to find the side length of each small square, we just have to square 64, and the answer is 8. Therefore, the side length of each small square is 8feet. Hope this helps.
Step2247 [10]3 years ago
3 0

Answer:It is 8 ft on edg

Step-by-step explanation:

You might be interested in
In a class of 7, there 4 syudents who play soccer. If the teacher chooses 3 students, what is the probability that none of the t
Lubov Fominskaja [6]

Answer:

75%

Step-by-step explanation:

Do 3/4 multiplied by 100% (100/100)

Then you get 75/100 which is 75%.

3 0
3 years ago
Ill give brainliest pls help i am so stuck
jok3333 [9.3K]

Answer:

Part A:

Since the slopes of the lines are not equal, the pair of equations for lines A and B will have one solution.

Part B:

(3 ,5 ) is the solution to the equations of lines A and B.

Step-by-step explanation:

Part A:

We are required to determine the number of solutions that the pair of equations for lines A and B have. To do this, we shall determine the slope of each line;

A straight line labeled A joins the ordered pair 2, 6 with the ordered pair 6, 2. The slope is defined as;

(change in y) / (change in x)

The slope of line A is thus;

(2 - 6)/(6 - 2) = -1

A straight line B joins the ordered pair 0, 3 with the ordered pair 4.5, 6.

On the other hand, the slope of line B is;

(6 - 3) / (4.5 - 0) = 2/3

Since the slopes of the lines are not equal, the pair of equations for lines A and B will have one solution.

Part B:

What is the solution to the equations of lines A and B?

We first need to determine the equation of each line;

Line A, the slope was found to be -1 and the line passes through (2,6)

The equation of the line in slope intercept form is;

y = -x + c

6 = -2 + c

c = 8

y = -x + 8

On the other hand, the slope of line B was found to be 2/3 and the line passes through (0,3). Therefore, we have;

y = 2/3x + c

3 = 2/3 (0) + c

c= 3

y = (2/3)x + 3

The graph of these two lines is shown in the attachment below. The lines intersect at (3 ,5 ) which is the solution to the equations of lines A and B.

The graphical solution to a system of linear equations is given by the point of intersection of the two lines representing each equation.

3 0
3 years ago
Imagine a world in which only decimal,not fractions,are used.how would your life be different
Firdavs [7]
This answer depends a bit on your age, the types of activities you partake in and the kind of work you do/are planning to do but here goes:

I am thinking of some uses of fractions where decimals are not typically used. One might be cooking. Often the ingredients (1/2 cup of four and so on) are measured using fractions. If you were in a world with decimals you might need to make (1/3) the servings of a recipe that calls for 1/4 of a cup of some ingredient and instead of 1/12 have to deal with a long repeating decimal that probably would need to be approximated so would not be precise.

While on the subject of food ordering pizza (1/2 with pepperoni, 1/4 mushrooms and 1/4 plain) would be doable after you got used to it but probably not as comfortable. Dividing up slices of pizza among friends (one slice is usually 1/8 of a pie) might be awkward though eventually doable.

Estimation - the biggest issue is exactitude versus estimation. When we use a fraction like 1/3 that is an exact value, but when we use .333 or .3333333 no matter how many 3s we use we are only estimating because the 3s go on forever and we can't write them forever. Yes, we can use .3 (with a bar over the 3, but now try to multiply that with .456565656 with a bar over the 56. This becomes practically impossible unless we estimate ... so the biggest issue would be that you would lose precision in many calculations and measurements and have to deal with answers that are good enough (but not exact).

Now say you work on some major car company or you design bridges or you are a scientist developing medicine that cures diseases, would not you want the ability to measure and compute precisely? If I split the pizza up wrong it is not a big deal. If I use a little more flour or a little less than I should in the recipe it might not make much of a difference in the end but if I am doing something that impacts the health, safety or well being of another human being, I would not want to live in a world where I have to estimate and can't count on having the exact, precise value.
3 0
3 years ago
The diameter of fishing line varies. Fishing lines can have a diameter as small as 2 x 102 inch and as large as 6 x 10-2 inch. H
Reil [10]
Answer:
3 times

Explanation:
We know that:
small diameter = 2 * 10^-2 in
large diameter = 6 * 10^-2 in

We want to know how many times larger is the thin diameter compared to the large one.
We will do this as follows:
large diameter = k * small diameter
where k is the number of times that we want to find
6 * 10^-2 = k * 2 * 10^-2
k = (6 * 10^-2) / (2 * 10^-2)
k = 3

This means that the large diameter is 3 times the small one.

Hope this helps :)
4 0
3 years ago
Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
Scorpion4ik [409]

let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

4 0
3 years ago
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