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3 years ago

How do I state the excluded values for this rational expression: 60x^3/12x

Mathematics

1 answer:

Stella [2.4K]3 years ago

7 0

Answer:

5x^2

Step-by-step explanation:

12 can be taken out from the top and bottom then for the x you subtract

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How do we figure this out

lisov135 [29]

Answer:

Pair 1: 10 cm and 2 cm. and the sum is 12 cm.

Pair 2: 20 cm and 4 cm. and the sum is 24 cm.

Step-by-step explanation:

Area of a rectangle is given by A = LW, where L is the length and W is the width of the rectangle.

Pair 1:

For the first rectangle W = 4 cm and A = 40 sq, cm

Then $L = \frac{A}{W} = \frac{40}{4} = 10$ cm.

For the second rectangle W = 4 cm and A = 8 sq, cm

Then $L = \frac{A}{W} = \frac{8}{4} = 2$ cm.

Therefore, the sum of two unknown lengths = 10 + 2 = 12 cm. (Answer)

Pair 2:

For the first rectangle W = 4 cm and A = 80 sq, cm

Then $L = \frac{A}{W} = \frac{80}{4} = 20$ cm.

For the second rectangle W = 4 cm and A = 16 sq, cm

Then $L = \frac{A}{W} = \frac{16}{4} = 4$ cm.

Therefore, the sum of two unknown lengths = 20 + 4 = 24 cm. (Answer)

8 0

3 years ago

412.638 in expanded form

ser-zykov [4K]

(4x100)+(1x10)+(2x1)+(6x1/10)+(3x1/100)+(8x1/1000)

5 0

3 years ago

Read 2 more answers

BRAINLIEST AWARD NO.2

zzz [600]

Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

Step-by-step explanation:

The formula is given as recursive since it involves the previous number of the sequence, and defined as:

$a_n=a_{n-1}*3+6$

we also know that the first term is 4

Then in this case, the first five terms are:

$a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\$

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

$a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2$

So the rule is to subtract 6 from he term, and divide the subtraction by 3. Then working backwards:

$a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49$