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3 years ago

Match the expressions given in words with their values when m=6.

Mathematics

1 answer:

Gemiola [76]3 years ago

4 0

Answer:

Value of expression 1 = 3

Value of expression 2 = 42

Value of expression 3 = 15

Value of expression 4 = 2

Step-by-step explanation:

Expression 1 : the sum of m and 3 divided by the difference of m minus 3

Expression 1 $: \frac{m+3}{m-3}$

Substitute m = 6

So, value of expression 1= $\frac{6+3}{6-3}=3$

Expression 2 : the sum of 3 times m and 4 times m

Expression 2: 3m+4m

Substitute m = 6

So, value of expression 2=3m+4m=3(6)+4(6)=42

Expression 3 : the difference of the product of 3 and m minus the quotient of m divided by 2

Expression 3: $3m-\frac{m}{2}$

Substitute m = 6

So, value of expression 2= $3m-\frac{m}{2}=3(6)-\frac{6}{2}=15$

Expression 4 :the quotient of 6 divided by the difference of m minus 3

Expression 4: $\frac{6}{m-3}$

Substitute m = 6

So, value of expression 4= $\frac{6}{m-3}=\frac{6}{6-3}=2$

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Prove that under root 2 is a irrational number

torisob [31]

Answer:

Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2=(2k)2/b22=4k2/b22*b2=4k2b2=2k2

This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

(I copied this from the internet, but hope it helps!)

4 0

3 years ago

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DO NOT ANSWER THIS QUESTION FOR THE NEXT ONE MINUTE PLEASE PLEASE DONT

Advocard [28]

Answer:

who is this

Step-by-step explanation:

who deleated this and why

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2 years ago

What is lim x→-3 sqrt x^2-8

vitfil [10]

If the -8 is under the square root, then...

$\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\$

If the -8 is not under the square root, then...

$\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5$

Either way, we replace x with -3 and simplify.

For more information, refer to the direct substitution rule for limits.

4 0

2 years ago

What is the value of x? <br>enter your answer in the box.<br>x = _

UNO [17]

Answer: x = 7

Step-by-step explanation:

24 = 2x + 10

24 - 10 = 2x + 10 - 10 (Subtract 10 from both sides)

14 = 2x

14/2 = 2x/2 (Divide both sides by 2)

7 = x

x = 7

8 0

3 years ago

BRAINLIEST!!Felicia had $80 when she went to a music store that charges $18 for each CD. She needs to have no less than $20 in h

xxMikexx [17]

Answer:

The possible number of CDs she could buy is 1, 2, and 3.

Step-by-step explanation:

First, you have to make an equation to solve to find the answer(s):

80 - (18 · x) ≥ 20
80 is how much money Felicia has; 18 is for the cost of each CD; x is for the number of CDs; ≥ is no less; and 20 is how much money Felicia needs to have left.

1.) 80 - (18 · 1) ≥ 20

80 - 18 · 1 ≥ 20

80 - 18 ≥ 20

62 ≥ 20

Since 62 ≥ 20 is always true, there are infinitely many solutions.

2.) 80 - (18 · 2) ≥ 20

80 - 18 · 2 ≥ 20

80 - 36 ≥ 20

44 ≥ 20

Since 44 ≥ 20 is always true, there are infinitely many solutions.

3.) 80 - (18 · 3) ≥ 20

80 - 18 · 3 ≥ 20

80 - 54 ≥ 20

26 ≥ 20

Since 26 ≥ 20 is always true, there are infinitely many solutions.

4.) 80 - (18 · 4) ≥ 20

80 - 18 · 4 ≥ 20

80 - 72 ≥ 20

8 ≥ 20

Since 8 ≥ 20 is false, there is no solution.

5.) 80 - (18 · 5) ≥ 20

80 - 18 · 5 ≥ 20

80 - 90 ≥ 20

- 10 ≥ 20

Since - 10 ≥ 20 is false, there is no solution.

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4 years ago

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