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3 years ago

Is negative two less than negative one?

Mathematics

2 answers:

Juliette [100K]3 years ago

5 0

Answer:

Yes negative two is less that negative one. The farther you move to te left of zero the smaller the number becomes.

geniusboy [140]3 years ago

3 0

Answer:

Yes.

Step-by-step explanation: The confusing thing about negetives is the greater the negetive number is, the less the value is. think of a number line. the farther the number is to the left, the smaller it is.

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What is the mean of the number 48,78,54,68

lorasvet [3.4K]

Answer:

Step-by-step explanation:

$mean = \frac{48 + 78 + 54 + 68}{4} \\ = \frac{248}{4} \\ = 62$

6 0

3 years ago

a cell phone plan offers 250 minutes each month for 15.00 dollars per month. what is the cost per minute when a person uses all

Nezavi [6.7K]

It costs 0.06 cents each minute.

Explanation : $15/250 minutes = $0.06

4 0

3 years ago

Staples sell for $60 a box .How many boxes can be bought for 3.45

kakasveta [241]

Answer:0.0575

Step-by-step explanation:

1 box cost $60

x boxes can be bought for $3.45

x=3.45/60

x=0.0575

4 0

3 years ago

The sum of three numbers is 10. Two times the second number minus the first number is equal to 12.

eimsori [14]

Answer:

-10,1,19

Step-by-step explanation:

x+y+z = 10 (Equation 1)

2y-x= 12 (Equation 2)

x-y+2z = 7 (Equation 3)

(Equation 2): -x = -2y+12

x = 2y-12 (Equation 4)

(Equation 1) - (Equation 3): 2y-z = 3

-z = -2y+3

z = 2y-3 (Equation 5)

Substitute (4) and (5) into (1)

x+y+z = 10

(2y-12)+y+(2y-3) = 10

5y-15 = 10

5y = 5

y=1

Substitute y=1 into (2)

2y-x= 12

2(1)-x= 12

2-x= 12

-x= 12-2

-x= 10

x= -10

Substitute y=1 and x=-10 into (1)

x+y+z = 10

-10+1+z = 10

z-9 = 10

z = 10+9

z = 19

Order: x = -10, y = 1, and z = 19

8 0

3 years ago

Read 2 more answers

Prove that for all n in the naturals, n(n+1)(n+2) is divisible by 3

DerKrebs [107]

By induction:

It's true for $n=1$ , since $1\cdot2\cdot3$ clearly contains a factor of 3.

Suppose it's true for $n=k$ , that $k(k+1)(k+2)$ is divisible by 3. Then

$(k+1)(k+2)(k+3)=\dfrac{k(k+1)(k+2)(k+3)}k=\dfrac{3m(k+3)}k$

where $m$ is an integer. This reduces to

$\dfrac{3m(k+3)}k=3m+9\dfrac mk$

and both terms are clearly multiples of 3. We know that $\dfrac mk$ is an integer since we had set $m=k(k+1)(k+2)$ previously, which implies $m$ is a multiple of $k$ . So the statement is true.

6 0

3 years ago