What is the two ratios equivalent to 14:32

Draw at least six different sized rectangles that have an area of 64 square units

olga_2 [115]

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

$A=x*y$

$A=64\ units^{2}$

so

$x*y=64$ --------> equation 1

let's assume different values of x to get the different values of y

case 1) For x=64 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/64$

$y=1\ units$

the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

case 2) For x=32 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/32$

$y=2\ units$

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

case 3) For x=16 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/16$

$y=4\ units$

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

case 4) For x=30 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/30$

$y=\frac{32}{15} =2\frac{2}{15} \ units$

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

case 5) For x=40 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/40$

$y=1.60\ units$

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

case 6) For x=60 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/60$

$y=\frac{16}{15} =1\frac{1}{15} \ units$

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

3 0

3 years ago

Prime numbers problem :)

Hatshy [7]

Answer:

The answer to your question is (2)(3)(11)

Step-by-step explanation:

Data

154

Process

1.- Find the prime factors of 154, starting with 2, then, 3, 5, 7, etc

154 2

77 7

11 11

1

2.- Write 154 as a composition of prime factors

154 = (2)(7)(11)

3.- Conclusion

The prime factors of 154 are 2, 3 and 11

7 0

3 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

Does anyone know all the concepts/rules needed to learn math in college? I want to make sure I have things I need now in my seni

Mandarinka [93]

Answer:

Basic college math topics include whole numbers, fractions, numbers, decimals and integers. Problem solving, algebra, percents and geometry are also included in course content.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Can somebody please help me with #11 thank you

aniked [119]

The graph is not a function, as it does not pass the vertical line test. The lines at 2,-1 and 2,3 overlay and pass the vertical line more than once, meaning that the graph is not a function.

6 0

3 years ago