1)x+30=40
x=40-30
Answer :x=10
2)30-20+2x=10
10+2x=10
2x=0
Answer: x=0
3)3x-10+13=-2x+28
3x+3=-2x+28
3x+2x+3=28
3x+2x=28-3
5x=28-3
5x=25
x=5
4)13x-23-45=-7x+12
13x-68=-7x+12
13x+7x-68=12
13x+7x=12+68
20x=12+68
20x=80
x=4
5)2(x+4)=10x+24
2x+8=10x+24
2x-10x+8=24
2x-10x=24-8
-8x=24-8
-8x=16
x=-2
6)3(x-5)=1-(2x-4)
3x-15=1-(2x-4)
3x-15=1-2x+4
3x-15=5-2x
3x+2x-15=5
3x+2x=5+15
5x=5+15
5x=20
x=4
7)5x-10=15
5x=15+10
5x=25
x=5
8)x+4=x+6
4=6
no solution
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What is the nth term to this sequence 18,16,14,12
?
Answer: 4
Explanation:
1. 18
2. 16
3. 14
4. 14
5. 12
6. 10
7. 8
8. 6
9. 4
Answer: d
Step-by-step explanation:
Answer:
= 4n - 2
Step-by-step explanation:
There is a common difference d between consecutive terms in the sequence, that is
6 - 2 = 10 - 6 = 14 - 10 = 18 - 14 = 4
This indicates the sequence is arithmetic with n th term
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 2 and d = 4 , thus
= 2 + 4(n - 1) = 2 + 4n - 4 = 4n - 2
Answer:
144 feets
Step-by-step explanation:
Look on s(t) like quadratic function. We need to find maximum of the function using 1st diferetioal:
s(t)=-16t^2+64t+80
s’(t)=-32t+64
S’(t)=0 if t=2
For t>2, s’(t)<0
For t<2, s’(t)>0. So the maximum od function is in the point 2.
S(2)=-16*4+64*2+80=-64+128+80=144
