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7nadin3 [17]
3 years ago
14

Ab-0.5bab−0.5ba, b, minus, 0, point, 5, b when a=1a=1a, equals, 1 and b=5b=5

Mathematics
1 answer:
barxatty [35]3 years ago
4 0

Answer:

  2.5

Step-by-step explanation:

Put the values in place of the corresponding variables and do the arithmetic:

  ab - 0.5b = (1)(5) -0.5(5) = 5 - 2.5 = 2.5

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1.2x=6 what is the answer
kifflom [539]

Answer:

To figure this out we need to balance out the equation.

When a number is right next a variable, it is considered as multiplication. So, to balance the equation we need to divide.

Step-by-step explanation:

1.2/1.2 =x

1.2/6=5

Answer; x=5

7 0
3 years ago
He first three terms of a geometric sequence are as follows. 10 , 30 , 90 find the next two terms of this sequence.+
ankoles [38]
10*3=30
30*3=90
90*3=270
270*3=810

or

10*3^1=30
10*3^2=90
10*3^3=270
10*3^4=810

the next two terms of this sequence are 270 and 810.
4 0
3 years ago
17+(−8)+(−8) help plweasee
madam [21]

Answer:

1

Step-by-step explanation:

17+(-8)+(-8)

9+(-8)

1

6 0
4 years ago
Read 2 more answers
PLEASE HELP,I AM BEING TIMED!!
Yuri [45]

Answer:

<h2><em>angle</em><em> </em><em>of</em><em> </em><em>1</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>6</em><em>0</em><em>°</em></h2>

Step-by-step explanation:

<h2 /><h2>we know vertical opposite angle are equall so </h2><h2><em>angle</em><em> </em><em>1</em><em> </em><em>=</em><em> </em><em>6</em><em>0</em><em>°</em></h2>
7 0
3 years ago
Write the first three terms of the series for which tn = 2(n+3). Find the number of terms of the series required to make the sum
ira [324]

Answer:

The first three terms of the series are 8, 10 and 12. The number of terms is 12 to make the sum 228.

Step-by-step explanation:

The series is defined as

t_n=2(n+3)

Put n=1.

t_1=2(1+3)=2\times 4=8

Put n=2.

t_2=2(2+3)=2\times 5=10

Put n=3.

t_3=2(3+3)=2\times 6=12

The first three terms of the series are 8, 10 and 12.

It is an arithmetic series. The first terms is 8 and the common difference is

d=a_2-a_1=10-8=2

The sum of n terms of an arithmetic series is

S_n=\frac{n}{2}[2a+(n-1)d]

288=\frac{n}{2}[2(8)+(n-1)2]

288=\frac{2n}{2}[8+n-1]

288=n[n+7]

0=n^2+7n-288

0=n^2+19n-12n-288

0=n(n+19)-12(n+19)

0=(n+19)(n-12)

Equate each factor equal to zero.

n=-19,12

The number of terms can not be negative, therefore the value of n must be 12.

6 0
3 years ago
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