Lesson 2: Multiplying and Dividing Rational Expressions

max is making a rectangular garden that is 5 feet less than twice its width. If the perimeter of the garden is 80 feet, what wil

djverab [1.8K]

L=2w-5

2(l+w)=80
l+w=40
2w-5+w=40
3w=45
w=15 feet
l=25 feet. As a result, the width is 15 feet and length is 25 feet. Hope it help!

7 0

3 years ago

Read 2 more answers

Solve the simultaneous equation, 2p - 3q = 4 3p + 2q = 9

Kobotan [32]

Answer:

p = $\frac{35}{13}$ and q = $\frac{6}{13}$

Step-by-step explanation:

Given equations:

2p - 3q = 4 -----------(i)

3p + 2q = 9 ------------(ii)

Let's solve this equation simultaneously using the elimination method

(a) Multiply equation (i) by 3 and equation (ii) by 2 as follows;

[2p - 3q = 4] x 3

[3p + 2q = 9] x 2

6p - 9q = 12 -------------(iii)

6p + 4q = 18 -------------(iv)

(b) Next, subtract equation (iv) from equation (iii) as follows;

[6p - 9q = 12]

- [6p + 4q = 18]

-13q = -6 -----------------(v)

(c) Next, make q subject of the formula in equation (v)

q = $\frac{6}{13}$

(d) Now substitute the value of q = $\frac{6}{13}$ into equation (i) as follows;

2p - 3( $\frac{6}{13}$ ) = 4

(e) Now, solve for p in d above

Multiply through by 13;

26p - 18 = 52

Collect like terms

26p = 52 + 18

26p = 70

Divide both sides by 2

13p = 35

p = $\frac{35}{13}$

Therefore, p = $\frac{35}{13}$ and q = $\frac{6}{13}$

8 0

2 years ago

Plzzzzzzzz! A suitcase lock has a code 4 symbols long. The symbols on the lock are a star, a square, a circle, a triangle, a lin

34kurt

Answer:

$\dfrac{1}{30}$

Step-by-step explanation:

6 symbols and 4-symbol code so number of combinations wich doesn't contain the same symbol twice would be:

6•5•4•3 = 360

If code starts with a star followed by triangle that means only one posibility for first two symbols and code doesn't containing the same symbol twice means that left 4 symbols for next two.

So number of combinations of code starting with a star followed by triangle and doesn't containing the same symbol twice will be:

1•1•4•3 = 12

The probability that a randomly chosen code starts with a star followed by triangle and doesn't contain the same symbol twice:

$\dfrac{1\cdot1\cdot4\cdot3}{6\cdot5\cdot4\cdot3}=\dfrac{1\cdot1}{6\cdot5}=\dfrac{1}{30}$

7 0

3 years ago

For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7

Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be approaching to a value $\infty$ or it should not give us the $\frac{0}{0}$ (zero by zero) form when the input is given to the function.