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sergeinik [125]
3 years ago
7

Urgent!! Answer like this. “3rd graph goes to the 4th item drop” please or however you want :)

Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

graph 1 = greater than 1

graph 2 = 0

graph 3 = 0

graph 4 = less than 0

graph 5 - between 0 & 1

Sorry if anything is wrong!!

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the length of a rectangle is 5 inches more then the width. the area of the rectangle is equal to 2 inches more than 4 times the
Svet_ta [14]
P = 2(L + W)
L = W + 5
A = 4P + 2

P = 2(W + 5 + W)
P = 2(2W + 5)
P = 4W + 10

A = 4P + 2
A = 4(4W + 10) + 2
A = 16W + 42

A = L * W
A = W(W + 5)
A = W^2 + 5W

W^2 + 5W = 16W + 42
W^2 + 5W - 16W - 42 = 0
W^2 - 11W - 42 = 0
(W + 3)(W - 14) = 0

W - 14 = 0
W = 14 <==

L = W + 5
L = 14 + 5
L = 19 <==

P = 2(19 + 14)
P = 2(33)
P = 66

A = L * W
A = 19 * 14
A = 266

answer : length = 19, width = 14....perimeter = 66....area = 266
5 0
3 years ago
Which system of equations has the same solution as the system below?
worty [1.4K]

Answer: Both A, and C

Step-by-step explanation:

The answer to the first system of equations (2x+2y=16) would be

x=3 and y=5 ( 3x-y=4 )

Which means we have to find out which of the other equations has an x value of 3, and a y value of 5.

If A is 2x+2y=16, then x=3 and y=5

           6x-2y=8

If B is x+y=16, then x=5 and y=11

         3x-y=4

If C is 2x+2y=16, then x=3 and y=5

           6x-2y=8

If D is 6x+6y=48 , then x=-2 and y=10

            6x+2y=8

Both A and C are equal to the first system of equations, which means they are both correct answers.

8 0
3 years ago
Factor the expression below.
Genrish500 [490]

Answer:

A. (6x + 7)(6x - 7)

Step-by-step explanation:

use difference of squares, which says a^2 - b^2 =(a+b)(a-b)

take the square roots of the two numbers, 6x and 7, and use them as a and b

8 0
2 years ago
Help! anyone SMART PLEASE
bagirrra123 [75]
Answer choice B is the correct answer the rest don't make sense.
5 0
3 years ago
X2 + 3x - 6 = 0
Vesnalui [34]

Answer:

Go to ma th way Step-by-step explanation:

6 0
3 years ago
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