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3 years ago

12

two garden plots are to have the same area. one is a square and one is rectangular. the rectangular plot is 4 meters wide and 16

meters long. how large is one side of the square garden plot in meters?

1 answer:

kirza4 [7]3 years ago

5 0

In this question, there are two garden with same area but different shape (rectangle and square). The formula for rectangle area is length* width, so the area of the garden would be:

rectangle garden area= 4m * 16m= 64m^2

The formula for the square area would be side^2. The square garden area is same as the rectangle garden area. Combining both equations will result in:

square garden area= rectangle garden area
side^2= 64m^2
side= √64m^2= 8m

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What is the formula for the area A of a trapezoid with parallel sides of length B and D, nonparallel sides of length A and C and

loris [4]

Answer:

$(B) \dfrac12H (B+D)$

Step-by-step explanation:

$\text{Area of a trapezoid }= \dfrac12 ($Sum of the parallel sides) \times $Height\\Parallel Sides = B and D\\Height =H\\Therefore:\\\text{Area of the trapezoid }= \dfrac12 (B+D) H$

The correct option is B.

4 0

3 years ago

What is the sum _71/6 +35/6 in the simplest terms ? Show your work

adoni [48]

Answer:

The answer in simplest term is 35.33333334.

Step-by-step explanation:

First, we need to divide 71 and 6.

71/6 = 23.66666667.

Now we need to divide 35 and 6.

35/6 = 11.66666667.

Now we need to add the two quotients.

23.66666667 + 11.66666667 = 35.33333334.

And that is the final answer.

I hope this helps!

7 0

2 years ago

Property being illustrated 73=0+73

Bad White [126]

Answer:

yes???

Step-by-step explanation:

your question either doesn't make sense or I just don't know

8 0

2 years ago

With a little care, you can find the median and the quartiles from the histogram. what are these numbers? you can also find the

coldgirl [10]

Complete Question

The histogram from this question is shown on the first uploaded image

Answer:

The first quartile is $1st Q = 19^{th} girl (subject ) = 1 \serving\ of\ fruit\ per\ day$

The median is $Median = 38 ^{th} \ girl (subject) = 2 \serving\ of\ fruit\ per\ day$

The third quartile is $3rd Q = 56^{th} \ girl (subject) = 4 \serving\ of\ fruit\ per\ day$

The mean is $\= x = 2.62$

Step-by-step explanation:

From the question we are told that

The total number of girls is n = 74

Generally the median is mathematically represented as

$Median = \frac{\frac{n}{2} +[ \frac{n}{2} + 1] }{2}$

So

$Median = \frac{\frac{74}{2} +[ \frac{74}{2} + 1] }{2}$

=> $Median = \frac{37 +38 }{2}$

=> $Median =37.5 \ girl$

From the histogram $37.5 \approx 38^{th} \ girl$ fall under 2 fruits per day

Generally the first quartile is mathematically represented as

$1st \ Q = \frac{\frac{n}{4} + [\frac{n}{4} + 1] }{2}$

=> $1st \ Q = \frac{\frac{74}{4} + [\frac{74}{4} + 1] }{2}$

=> $1st \ Q = 19 \ girl$

From the histogram $19^{th}$ girl fall under 1 fruit per day

Generally the third quartile is mathematically represented as

$3rd\ Q = \frac{\frac{n}{2} + n }{2}$

=> $3rd\ Q = \frac{\frac{74}{2} + 74 }{2}$

=> $3rd\ Q = 55.5$

From the histogram $55.5 \approx 56^{th} \ girl$ fall under 4 fruits per day

Generally from the histogram table the frequency [number of subjects (girls) ] and the servings of fruit per day can be represented as

Total

x(servings per day ) 0 1 2 3 4 5 6 7 8

f (number of subjects ) 15 11 15 11 8 5 3 3 3 $\sum f = 74$

xf 0 11 30 33 32 25 18 21 24 $\sum xf = 194$

Generally the mean is mathematically represented as

$\= x = \frac{1}{ \sum f} * [\sum xf]$

=> $\= x = \frac{1}{ 74} *194$

=> $\= x = 2.62$

4 0

2 years ago

If a square has a side of 4', what would the area be?<br>A= s2=sxs<br>A= 42= 4'x4<br>A= 16 sq. ft.

statuscvo [17]

Answer:

16 sqft.

Step-by-step explanation:

A=L(W)

A=4(4)

A=16

Hope that helps

4 0

2 years ago