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3 years ago

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Rectangles A and B have the same perimeter but are different shapes Rectangle S is 5 inches long and 3 inches wide. Rectangle B

is 6 inches longer than it is wide. What are the lenght and width of Rectangle B ?

1 answer:

koban [17]3 years ago

4 0

We know that
perimeter rectangle A=2*[5+3]-----> 16 in

let
x------> length side rectangle B
y-------> width side rectangle B
so
x=6+y-------> equation 1
16=2[x+y]------> equation 2
<span>I substitute 1 in 2
</span>16=2*[6+y+y]-------> 16=2*[6+2y]-----> 16=12+4y-----> 4y=16-12
y=4/4--------> y=1 in
x=6+1---->x=7 in

the answer is
the dimensions of rectangle B are
length side is 7 in
width side is 1 in

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Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

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2 years ago

What is the probability that a two-digit number selected at random has a tens digit less than its units digit

Reika [66]

The probability that a two-digit number selected at random has a tens digit less than its units digit is 0.2667 (4/15).

$A=24B=90\\\\P=A/B\\\\P=24/90\\p=4/15$

There are 90 two-digit numbers (99-9). Of these, six numbers are divisible by 15 (15, 30, 45, 60, 75, 90). This is also divisible by 5. Therefore, the preferred case is 30-6 = 24. Therefore, the required probability is 24/90 = 4/15.

The probability of an event can be calculated by simply dividing the number of favorable results by the total number of possible results using a probabilistic expression. Whenever you are uncertain about the outcome of an event, you can talk about the probability of a particular outcome, that is, its potential.

Learn more about probability here: brainly.com/question/24756209

#SPJ4

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1 year ago

What is the output value of 2x^2-9x-5 input value is x=2

Kobotan [32]

Answer:

-15

Step-by-step explanation:

Go to (Mâthway) This page can help you with problems like that. (replace the â with a when searching the site.)

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Step-by-step explanation:

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Step-by-step explanation:

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