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Black_prince [1.1K]
3 years ago
10

Which of the following inequalities have the same solution set? Select all that apply (>= means greater than or equal to and

<= means less than or equal to) *
0 points
7n + 4 - 5n <= 2(n+2)
7 + 6a >= -4(2-a) -2a
2(x+4) - 4x>= -8
7-6a>= -6a -7
-3(n-4) >= 10-3n
Mathematics
1 answer:
Alex_Xolod [135]3 years ago
5 0
<span>7n + 4 - 5n <= 2(n+2) Let's simplify these expression first 2n + 4 <= 2(n + 2) 2(n +4) = 2n + 4 The LHS and RHS expressions are the same but the inequality says contrary Hence 2n + 4 is not less than but equal to 2(n + 2). 7+ 6a >= -4(2-a) - 2a -4(2-a) - 2a = -8 + 4a - 2a = -8 + 2a. It should be obvious that 7 +6a is greater than the RHS expression. Suppose a= 2 then we have 7 + 6(2) = 19 while -8 + 2(2) = - 8 + 4 = -4 So it follows that 7 + 6a >= -4(2-4) -2a 4x >= 8. The expression is always true so the equal to option is out.</span>
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3 years ago
Which shows the given inequalities in slope-intercept form?
yanalaym [24]

Answer: Solve for X      

4x−5y<1

Step-by-step explanation:

step 1 add 5y to each side

4x−5y+5y<1+5y

4x<5y+1

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4x4<5y+14

x<54y+14

Answer:

x<54y+14

Let's solve for x.

12y−x<3

Step 1: Add (-1)/2y to both sides.

−x+12y+−12y<3+−12y

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Step 2: Divide both sides by -1.

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x>12y−3

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4 0
3 years ago
(Worth 30 points) What are the explicit equation and domain for an arithmetic sequence with a first term of 5 and a second term
kolbaska11 [484]

Answer:

a_n=5-2(n-1), all integers where n≥1

Step-by-step explanation:

we know that

The explicit equation for an arithmetic sequence is equal to

a_n=a_1+d(n-1)

a_n is the th term

a_1 is the first term

d is the common difference

n is the number of terms

we have

a_1=5\\a_2=3

Remember that

In an Arithmetic Sequence the difference between one term and the next is a constant, and this constant is called the common difference.

To find out the common difference subtract the first term from the second term

d=a_2-a_1=3-5=-2

substitute the given values in the formula

a_n=5-2(n-1)

The domain is all integers for n\geq 1

3 0
3 years ago
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