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atroni [7]
3 years ago
9

Solve for b 3(b-2) =2b+9

Mathematics
2 answers:
Cloud [144]3 years ago
6 0
First distribute the 3

3b-6 = 2b+9

Then add 6 to each side of the equation

3b = 2b+15

Next subtract 2b from each side of the equation

b = 15

Your answer is b = 15

I hope this helps!
jarptica [38.1K]3 years ago
5 0
3b - 6 = 2b + 9 
3b-2b=9+6
b=15
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Calculate the value for each determinant:
Dafna1 [17]

Answer:

See explanations below

Step-by-step explanation:

Example of how to find the determinant of a 2×2 matrices is as shown;

\left[\begin{array}{ccc}a&b\\c&d&\end{array}\right]

= (a*d)-(b*c)

= ad - bc

Applying this to solve the given questions

1) (3*4) - (-3*-1)

= 12 - (3)

= 9

2) (-12*5)-(-1*12)

= -60 - (-12)

= -60+12

= -48

3) (0*4)-(-1*17)

= 0-(-17)

= 0+17

= 17

4) (8*-1)-(-4*-1)

= -8-(4)

= -8-4

= -12

5) (2*-1)-(2*-1)

= -2-(-2)

= -2+2

= 0

6) 1(1) - 0(0)

= 1 - 0

= 1

4 0
3 years ago
Need help with ASAP!
Evgesh-ka [11]

Answer: 8y^6x^6

Step-by-step explanation:

Volume = l*w*h so multiply all the variable by <u>multiplying the numbers and adding the exponents</u> so

2y^2*4x^4y=8y^3x^4

then 8y^3x^4*x^2y^3=8y^6x^6

8 0
3 years ago
Read 2 more answers
Help me (10 points )
Ipatiy [6.2K]

Answer:x=10

Step-by-step explanation:x x2

8 0
3 years ago
Principal Philippi use a random sample of 20 student records to determine how far in miles students live from the school the res
nexus9112 [7]

Take 3 4 5 3 4 5 6 5 4 3 2 3 4 5 6 4 8 4 3 2 and add them all together

=83

Now you divide 83 by the number of terms in the sample set, which is 20.  

83/20 = 4.15

For rounding to the nearest tenth, look at the hundredths space, which is 5. Since it is 5 or higher, round the tenths up one.  

final answer= 4.2

6 0
3 years ago
Read 2 more answers
Write the first six terms of the geometric sequence with the first term 6 and common ratio 1/3
Alja [10]

Answer:

6, 2, 2/3, 2/9, 2/27, 2/81

Step-by-step explanation:

The nth term of a geometric progression is expressed as;

Tn  = ar^n-1

a is the first term

n is the number of terms

r is the common ratio

Given

a = 6

r = 1/3

when n = 1

T1 = 6(1/3)^1-1

T1 = 6(1/3)^0

T1 = 6

when n = 2

T2= 6(1/3)^2-1

T2= 6(1/3)^1

T2 = 2

when n = 3

T3 = 6(1/3)^3-1

T3= 6(1/3)^2

T3= 6 * 1/9

T3 = 2/3

when n = 4

T4 = 6(1/3)^4-1

T4= 6(1/3)^3

T4= 6 * 1/27

T4 = 2/9

when n = 5

T5 = 6(1/3)^5-1

T5= 6(1/3)^4

T5= 6 * 1/81

T5 = 2/27

when n = 6

T6 = 6(1/3)^6-1

T6= 6(1/3)^5

T6= 6 * 1/243

T6 = 2/81

Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81

6 0
3 years ago
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