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3 years ago

WHICH INQUALITY IS EQUAL TO -1 > M+4

Mathematics

1 answer:

PIT_PIT [208]3 years ago

5 0

Answer:

m<−5

Step-by-step explanation:

Step 1: Flip the equation.

m+4<−1

Step 2: Subtract 4 from both sides.

m+4−4<−1−4

m<−5

Hope this helped <3

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MA_775_DIABLO [31]

Answer:

option A

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Step-by-step explanation:

refer to the above attachment

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2 years ago

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vfiekz [6]

Answer: The fourth term is -102

----------------------------------------------

Explanation:

The term after the nth term is generated by this rule $a_{n+1} = -4(a_n) + 2$ which means that we first
Step 1) multiply the nth term ( $a_n$ ) by -4
Step 2) Add the result of step 1 to the value 2 to get the next term in the sequence

Let's follow those steps above to generate the first four terms

The first term is $a_1 = 2$ . In short, the first term is 2

The second term is...
$a_{n+1} = -4(a_n) + 2$
$a_{1+1} = -4(a_1) + 2$
$a_{2} = -4(2) + 2$
$a_{2} = -8 + 2$
$a_{2} = -6$
So the second term is -6

The third term is...
$a_{n+1} = -4(a_n) + 2$
$a_{2+1} = -4(a_2) + 2$
$a_{3} = -4(-6) + 2$
$a_{3} = 24 + 2$
$a_{3} = 26$
The third term is 26

Finally, the fourth term is...
$a_{n+1} = -4(a_n) + 2$
$a_{3+1} = -4(a_3) + 2$
$a_{4} = -4(26) + 2$
$a_{4} = -104 + 2$
$a_{4} = -102$
The fourth term is -102.

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3 years ago

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2 years ago

Read 2 more answers

the difference between the squares of two numbers is 3. Twice the square of the first number increased by the square of the seco

Maurinko [17]

Answer:

The number are $\sqrt{6}$ and $\sqrt{6}$ .

Step-by-step explanation:

Given as :

The difference between the squares of two number = 3

And , Twice the square of the first number increased by the square of the second number is 9

Now, Let The first number = F

And The second number = S

So , According to question

F² - S² = 3 .......A

2 F² - S² = 9 ........B

Now, solving the equations

( 2 F² - S² ) - ( F² - S² ) = 9 - 3

Or, ( 2 F² - F² ) + ( S² - S² ) = 6

Or, F² + 0 = 6

∴ F = $\sqrt{6}$

Putting the value of F in eq A

So, ( $\sqrt{6}$ )² - S² = 3

Or, 6 - 3 = S²

Or, S² = 3

∴ S = $\sqrt{3}$

So, The number are $\sqrt{6}$ , $\sqrt{6}$

Hence The number are $\sqrt{6}$ and $\sqrt{6}$ . Answer

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2 years ago