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nignag [31]
3 years ago
6

Inequality Match Up Lab

Mathematics
2 answers:
Delvig [45]3 years ago
7 0

Answer:

1. 2x+3<-3 = x<-3

You get this by subtracting 3 from both sides, which gives you -6. Then divide 2 from both sides.

2. -3x<-6 = x>2

You divide -3 from both sides. You must switch the sign because any time you multiply or divide a negative integer from both sides of an inequality the sign is switched.

3. -5x<5 = x>1

See number 2

4. 5-2x<11 = x>-3

Subtract 5 from both sides to get -2x<6. Then, divide both sides by -2 and switch the sign.

5. x+2<5 = x<3

Subtract 2 from both sides

6. 2-3x<5 = x>-1

Subtract 2 from both sides to get -3x<3. Then, divide -3 from both sides and switch the sign.

Artist 52 [7]3 years ago
5 0
1. 2x+3<-3 = x<-3
You get this by subtracting 3 from both sides, which gives you -6. Then divide 2 from both sides.

2. -3x<-6 = x>2
You divide -3 from both sides. You must switch the sign because any time you multiply or divide a negative integer from both sides of an inequality the sign is switched.

3. -5x<5 = x>1
See number 2

4. 5-2x<11 = x>-3
Subtract 5 from both sides to get -2x<6. Then, divide both sides by -2 and switch the sign.

5. x+2<5 = x<3
Subtract 2 from both sides

6. 2-3x<5 = x>-1
Subtract 2 from both sides to get -3x<3. Then, divide -3 from both sides and switch the sign.
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Answer:

r=4cm

d Diameter

cm

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d=8cm

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d=2r

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Step-by-step explanation:

6 0
3 years ago
2. Find the 20th term of an arithmetic sequence if its 6th term is 14 and 14th term is 6.
Fudgin [204]

Answer:

\sf t_{20}= 0

Step-by-step explanation:

<h3>Arithmetic sequence:</h3>

      \sf \boxed{\bf n^{th} \ term = a + (n-1)d}\\\\\text{Here, a is the first term ; d is the common difference }

6th term is 14 ⇒ \sf t_6 = 14

                a + (6 - 1)d = 14

                    a  +  5d = 14  --------------(I)

14th term is 6 ⇒\sf t_{14} = 6

             a + (14-1)d = 6

                  a + 13d = 6 ----------------(II)

Subtract equation (II) from equation(I)

        (I)          a + 5d = 14

        (II)         a + 13d = 6

                    <u>-    -          -</u>

                            -8d = 8

                               d  = 8 ÷(-8)      

                              \sf \boxed{\bf d= (-1)}

Plugin d = -1 in equation (I)

a + 5(-1) = 14

      a -5  = 14

             a = 14 + 5

             \sf \boxed{\bf a = 19}  

20th term:

 \sf t_{20}= 19 + 19*(-1)

       = 19 - 19

   \sf \boxed{\bf t_{20} = 0}

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