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4 years ago

13

EXPERTS/ACE ( and people that wanna actually help)

1 answer:

goblinko [34]4 years ago

4 0

Consistent means it has at least one solution.....inconsistent means there is no solution....coincident means it is the same line and there are infinite solutions.

ur 2 equations are the same line...therefore, ur answer is consistent and coincident

You might be interested in

What is 4/7 equivalent to

VashaNatasha [74]

4/7 can be equivalent to 8/14 (Multiplied top and bottom of fraction by 2)

It can also be equivalent to 12/21. (Multiplied by 3)

16/28.

And so on.

5 0

3 years ago

HELPPPPPPP please and include work

gladu [14]

You simply multiply the 3 numbers 12*4*16=768

7 0

4 years ago

Two numbers are each multiplied by themselves to give two new numbers. The difference between these two new numbers is less than

steposvetlana [31]

Answer:

4 or 5.

Step-by-step explanation:

<u>Let the numbers be represented by </u><u>x</u><u> and </u><u>y</u><u>.</u>

x * x = mc

y * y = zq

m - c < 10

z - q < 10

x - y = 1

x + y > 7

<u>Let's start from 0.</u>

0

squared: 0, <u>wrong; a new number is meant to be formed.</u>

1

squared: 1, <u>wrong; a new number is meant to be formed.</u>

2

squared: 4, <u>maybe this can work; a new number is formed.</u>

3

squared: 9, <u>maybe this can work; a new number is formed.</u>

4

squared: 16, <u>maybe this can work; a new number is formed.</u>

5

squared: 25, <u>maybe this can work; a new number is formed.</u>

6

squared: 36, <u>maybe this can work; a new number is formed.</u>

7

squared: 49, <u>maybe this can work; a new number is formed.</u>

8

squared: 64, <u>maybe this can work; a new number is formed.</u>

9

squared: 81, <u>maybe this can work; a new number is formed.</u>

<u />

<u><em>FROM HERE, 0 & 1 ARE OUT.</em></u>

<u><em>THE QUESTION STATES THAT THE DIFFERENCE BETWEEN THE TWO ORIGINAL NUMBERS WAS ONE.</em></u>

<u><em>NOW WE HAVE 2, 3, 4, 5, 6, 7, 8, & 9.</em></u>

<u>2 & 3.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

3^2 - 2^2

9 - 4 = 5; there's a possibility for this to work since the difference of their squares is less than 10.

<u>3 & 4.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

4^2 - 3^2

16 - 9 = 7; there's a possibility for this to work since the difference of their squares is less than 10.

<u>4 & 5.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

5^2 - 4^2

25 - 16 = 9; there's a possibility for this to work since the difference of their squares is less than 10.

<u>5 & 6.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

6^2 - 5^2

36 - 25 = 11, THIS IS WRONG.

<u>6 & 7.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

7^2 - 6^2

49 - 36 = 13, THIS IS WRONG.

<u>7 & 8.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

8^2 - 7^2

64 - 49 = 15, THIS IS WRONG.

<u>8 & 9.</u>

THE DIFFERENCE BETWEEN THEIR SQUARES SHOULD BE LESS THAN 10 IF THEY CAN WORK.

9^2 - 8^2

81 - 64 = 17, THIS IS WRONG

<u><em>THE POSSIBLE NUMBERS ARE 2&3, 3&4, AND 4&5.</em></u>

<u><em>THE QUESTION STATES THAT THE TWO ORIGINAL NUMBERS ADDED TOGETHER WAS MORE THAN 7.</em></u>

<u>2&3</u>

2 + 3 = 5, WRONG, THE ADDITION OF THE TWO ORIGINAL NUMBERS SHOULD BE MORE THAN 7 FOR THIS TO WORK.

<u>3&4</u>

3 + 4 = 7, WRONG, THE ADDITION OF THE TWO ORIGINAL NUMBERS SHOULD BE MORE THAN 7 FOR THIS TO WORK.

<u>4&5</u>

4 + 5 = 9, CORRECT, THE ADDITION OF THE TWO ORIGINAL NUMBERS IS MORE THAN 7.

Therefore, the possible numbers are 4 & 5. All that is left is for you to pick which of the two is the final answer if you have options.

Please, thank, rate 5 stars, give brainliest. Thank you.

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6 0

3 years ago

In a bag of m&m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&m

olasank [31]

There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
$_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\
=1140$

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
$_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20$

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
$P=
\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}$

The probability is 1/57.

4 0

3 years ago

What is the variation constant for an inverse squared relationship where we know x = –4 and y = 3?

Ymorist [56]

Answer:

It's B.

Step-by-step explanation:

y = k /x^2

3 = k / (-4)^2

3 * 16 = k

k = 48.

8 0

3 years ago