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2 years ago

Please help for Algebra II

Mathematics

1 answer:

fgiga [73]2 years ago

5 0

Answer:

I cant see ittt.... what the question

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One column of number consist of 61

iVinArrow [24]

6100 is the one column

6 0

3 years ago

49 points in seven games unit rate

-Dominant- [34]

Answer:

Step-by-step explanation:

1. Do 49÷7 to get the average of 7

5 0

3 years ago

Factor the expression below using the greatest common factor 6p3- 2p2 - 8p

miss Akunina [59]

Answer:

The Factor of expression using the greatest common factor is $2p(3p^2- p-4)$

Step-by-step explanation:

Consider the provide expression.

$6p^3- 2p^2 - 8p$

We need to Factor the expression using the greatest common factor.

First look at the coefficients of the variable.

The coefficients are the factor of 2.

Variable p is the greatest common factor in the provided expression.

$2p\times 3p^2- 2p\times p- 2p\times 4$

$2p(3p^2- p-4)$

Hence, the Factor of expression using the greatest common factor is $2p(3p^2- p-4)$ .

7 0

2 years ago

If a,b, and c are prime numbers, do (a+b)(a+b) and cc have a common factor that is greater than 1?

Tanya [424]

Answer:

1. Information is not sufficient

2. Information is not sufficient

Step-by-step explanation:

Question Modified

If a,b, and c are prime numbers, do (a+b) and c have a common factor that is greater than 1?

(1) a,b, and c are all different prime numbers

(2) c≠2

Explanation

1. Let's assume values of 1,3 and 5 to a, b, and c respectively

a + b = 1+3 = 4

4 and 5 do not have any common factor aside 1

Let's assume values of 1,3 and 2 to a,b and c respectively

a + b = 1+3 = 4

4 and 2 have a common factor of 2

So, the available information is not sufficient

Let's assume values of 2,7 and 3 to a b and c response

a + b = 2 + 7 = 9

9 and 3 have a common factor of 3

Also, let's assume values of 2,7 and 5 to a, b and c

a + b = 2 + 7 = 9

9 and 5 do not have a common factor

So, the available information is also not sufficient to answer the question.

3 0

3 years ago

1/x - 1/(x+1) ≤ 0 Please help quickly

Ymorist [56]

$\frac{1}{x}- \frac{1}{x+1}= \frac{(x+1)-x}{x(x+1)}= \frac{1}{x(x+1)} \leq 0$

Therefore x(x+1)≤0, and if two numbers multiply to a negative number, one of them has to be negative and the other has to be positive. In this case, because x+1>x, x≤0, and x+1≥0

Therefore -1≤x≤0, and x can take up the value of any real number in the interval [-1,0]

7 0

2 years ago