1/10 of 400000 is 10 times as much as

consuela's living room is a rectangle with an area of 360 square feet. the width of the living room is 5/8 its length. what is t

Aleks [24]

Area of Consuela's rectangular living room = 360 square feet
Let us assume the the length of the living room = x
Then
width of the living room = (5/8) * x
= (5x/8)
Then
Area of the rectangle = Length * Width
360 = x * (5x/8)
360 = 5x^2/8
2880 = 5x^2
x^2 = 2880/5
x^2 = 576
x^2 = (24)^2 inches
x = 24
Then
The length of the rectangle is = 24 inches
And
The Width of the rectangle is = 24 * (5/8) inches
= 3 * 5 inches
= 15 inches

3 0

3 years ago

Read 2 more answers

How do I solve inequalities with fractions on both sides?

KIM [24]

Please comment if you have any questions... :)

6 0

3 years ago

Solve i give 28 point

SIZIF [17.4K]

the number is how many years

5 0

3 years ago

Able, Ben and Cal each played a game able’s score was six times ben’s score, Cale score was a hird time of Able’s score write do

Feliz [49]

Answer:

6:1:2

Step-by-step explanation:

Let a = Able's score, b = Ben's score, and c = Cal's score.

Since

Able's score was 6 times Ben's score, that means a = 6b.

Cal's score was a third of Able's score, so that means c = a/3. And since a = 6b, that means c = 6b / 3 = 2b.

Thus, the ratio of Able's score to Ben's score to Cal's score, a:b:c, is 6:1:2, because c is twice as much as b and a is 6 times as much as b.

3 0

4 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$