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Tatiana [17]
2 years ago
12

-x=9 pls help I’m behind

Mathematics
1 answer:
olchik [2.2K]2 years ago
6 0

Answer: x = 10

Step-by-step explanation:

-x = 9

Whenever there's a negative number you always know there's a 1 in front of the x

-1x =9

+1 +1

equals x= 10

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HELP ASAP PLEASE HELPPPPPPP PLEASE
ella [17]

Answer:

5 times 3

Step-by-step explanation:

you do 5 time 3. it is 15

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3 years ago
The surface area of the figure is?
MissTica
The answer would be C
6 0
2 years ago
Use the tests for divisibility to determine which numbers divide evenly into the given number.
drek231 [11]

The numbers that can divide 648 evenly, using the divisibility tests, include <u>1, 2, 3, 4, 6, 8, 9, 12, and 18</u> and some multiples of any two numbers.

<h3>What is the divisibility test?</h3>

A divisibility test is performed to identify whether a number can be divided evenly by a fixed divisor without a remainder and without actually performing the division process.

<u>Divisibility by 1</u>: Every number is divisible by 1 = 648 (648/1).

<u>Divisibility by 2</u>: 648 is an even number and divisible by 2 = 324.

<u>Divisibility by 3</u>: Sum the digits (18). 18 is divisible by 3 = 6 (18/3).

<u>Divisibility by 4</u>: The last two digits (4 and 8) form a number (12) that is divisible by 4

<u>Divisibility by 6</u>:  It is divisible by 2 and by 3

<u>Divisibility by 8</u>: If the hundreds digit is even, the number formed by the last two digits must be divisible by 8.

<u>Divisibility by 9</u>: Sum the digits (18). 18 is divisible by 9

<u>Divisibility by 12</u>: 648 is divisible by 3 and by 4.

<u>Divisibility by 18</u>: 648 is divisible by 2 and by 9.

Learn more about the divisibility tests at brainly.com/question/24125354

#SPJ1

5 0
1 year ago
Suppose that $1800 is borrowed for three years at an interest rate of 2% per year, compounded continuously. Find the amount owed
Schach [20]

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5 0
2 years ago
Find the area of the rectangle with a length of (x^2-2) and a width of (2x^2-x+2)
Katyanochek1 [597]
Area=legnth times width

so multiply them together use distributive property
a(b+c)=ab+ac so
in this problem
(a+b)(c+d+e)=(a+b)(c)+(a+b)(d)+(a+b)(e)

so
x^2-2 times (2x^2-x+2)=(x^2)(2x^2-x+2)-(2)(2x^2-x+2)=(2x^4-x^3+2x^2)-(4x^2-2x+4)
add like terms
2x^4-x^3+(2x^2-4x^2)-2x+4
2x^4-x^3-2x^2-2x+4


5 0
3 years ago
Read 2 more answers
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