HELP NOW PLSSS enter your answer as mixed number and simpliflied, -4 1/2 x 2 1/2

We are interested in the dimensions of a certain rectangle. this rectangle has length twice the side of the square and width thr

Mila [183]

Area of square = s^2
Area of Rectangle = lw
l = 2s

w+3 = s
solve for w, w=s-3
Now we have l and w.
Plug both into area of rectangle formula so: (2s)(s-3)
Since both areas are equal set both equations equal to each other:
(2s)(s-3)=s^2
Now simplify
2s^2-6s=s^2
s^2-6s=0, Solve for s.
Factor polynomial. s(s-6)=0 , s can be equal to 0 or 6. HOWEVER, you cannot have a side length of 0 therefore the side length has to be 6.
Now plug in s for the length formula for the rectangle:
l = 2s so... l = 2(6) so length of rectangle = 12.
Now plug in s for the width formula for the rectangle:
w+3=s so... w+3=6 so width of rectangle = 3.
Now the dimensions of the rectangle are 12 by 3. 12 being length and 3 width.
To CHECK:
Find area of rectangle:
A=lw so A=3 times 12 so A=36
Find area of square:
We know the side is equal to 6 so
A=s^2 so 6^2 = 36
The areas are equal that verifies the answer of 12 by 3.

8 0

2 years ago

Tephen read 12 books, 20 magazines, and 17 newspaper articles last year. based on this data, what is a reasonable estimate of th

Basile [38]

THERE IS A 41% CHANCE HIS NEXT READING MATERIAL WILL BE A MAGAZINE

6 0

2 years ago

Read 2 more answers

Find the value of x, if angle x measures 102 degrees

Over [174]

The answer is B
Have a good day!

3 0

2 years ago

Read 2 more answers

Expand the following:<br> a) 3(x + 5)<br> b) 4(x-4)<br> c) 2(5x + 1)<br> d) 7(x-9y)

Tema [17]

Answer:

solution,

$a. \: 3(x + 5) \\ \: \: = 3 \times x + 3 \times 5 \\ \: \: = 3x + 15$

$b. \: 4(x - 4) \\ \: \: = 4 \times x - 4 \times 4 \\ \: \: = 4x - 16$

$c. \: 2(5x + 1) \\ \: = 2 \times 5x + 2 \times 1 \\ \: \: = 10x + 2$

$d . \: 7(x - 9y) \\ \: = 7 \times x - 7 \times 9y \\ \: \: = 7x - 63y$

Hope this helps..

Good luck on your assignment..

8 0

2 years ago

Read 2 more answers

Plss answer!! <br>MERRY Christmas!!! <br>Critical Question. AAAAA

liberstina [14]

Answer:

Step-by-step explanation:

abc = 1

We have to prove that,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

We take left hand side of the given equation and solve it,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

Since, abc = 1,

$\frac{1}{c}=ab$ and c = $\frac{1}{ab}$

By substituting these values in the expression,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}$

$=\frac{1+b+ab}{1+b+ab}$

$=1$

Which equal to the right hand side of the equation.

Hence, $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

5 0

2 years ago