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serg [7]
3 years ago
13

Which of the following is not a way to represent the solution of the inequality 2(x − 1) greater than or equal to −12?

Mathematics
1 answer:
lesya [120]3 years ago
5 0

Let us solve the inequality first


2(x-1)\ge -12


Dividing through by 2 gives


(x-1)\ge -6


Grouping like terms gives

x\ge -6+1

This implies that;

x\ge -5

This corresponds to option A in words

If we rewrite as

-5\le x

it will correspond to option C in words

Representing on diagram corresponds to option D (See attachment)


All are correct ways to represent the solution except B.

The correct answer is option B





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Can someone please help find the answer and explain how you got it please.
tresset_1 [31]

Answer:

5

Step-by-step explanation:

Hello!

We put in 2 for x and 6 for y

\frac{2(6)-2}{2}

Now we can solve this

\frac{12 - 2}{2}

\frac{10}{2} = 5

The answer is 5

Hope this helps!

4 0
3 years ago
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7 x 2/3 in simplest form
kozerog [31]

Answer:

4.67

Step-by-step explanation:

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8 0
2 years ago
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If the smaller of two numbers is one-half of the larger number and the sum of the two numbers is 81 , find the numbers.
zavuch27 [327]
Let smaller number be x/2 and the larger number x.

x/2 + x = 81
x/2 + x = 3x/2
3x/2 = 81
3x = 162
x= 54

The smaller number is 54/2= 27 and the larger number is 54.
4 0
2 years ago
6. Find the sum of whole number divisible by 3 and lies between 100 and 200.​
MatroZZZ [7]

Answer:

4950

Step-by-step explanation:

first find the smallest multiple of 3 thats >100, 3*33=99, so 3*34=102 is the smallest multiple of 3 >100.

next find the greatest multiple of 3 <200, 3*66=198 which is the greatest multiple of 3<200

34-66+1=number of terms=33. then find the average of the terms: 102+198/2=150. 150*33=4950. the answer should be 4950

3 0
3 years ago
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Using the letters in the word INNOVATIVE, find the number of permutations that can be formed using 4 letters at a time. Show you
lyudmila [28]

Answer:  1) 5040 and 2) 165

Step-by-step explanation:

1) Here total number of letters = 10

The number of permutations that can be formed using 4 letters at a time

= P (10, 4)

= 10_P_4

= \frac{10!}{(10-4)!}

=  \frac{10!}{6!}

= \frac{10\times 9\times 8\times 7\times 6!}{6!}

= 10 × 9 × 8 × 7

= 5040

2) Here the total number of machine = 11

The different combinations of machines can Geoff choose from to use

= 11_C_3

= \frac{11!}{(11-3)!3!}

= \frac{11!}{8!3!}

= \frac{11\times 10\times 9\times 8!}{8!\times 6}

= 11 × 5 × 3

= 165

5 0
3 years ago
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