Answer:
a=8
Step-by-step explanation:
3a + 2 - 5a = -14
Combine like terms
-2a +2 = -14
Subtract 2 from each side
-2a +2-2 = -14-2
-2a = -16
Divide each side by -2
-2a/-2 = -16/-2
a = 8
Answer:
a = 8
Step-by-step explanation:
-2a + 2 = -14
-2a = -16
-16/-2 = 8
a = 8
t - 6(-2t + 1)
The correct expression which is equivalent to this expression t - 6(-2t + 1) is,
⇒ 13t - 6
Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.
We have to given that;
The expression is,
⇒ t - 6 (- 2t + 1)
Now, We can simplify as;
⇒ t - 6 (- 2t + 1)
⇒ t + 12t - 6
⇒ 13t - 6
Thus, The correct expression which is equivalent to this expression
t - 6(-2t + 1) is,
⇒ 13t - 6
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Answer:
50th term is 159
Step-by-step explanation:
nth term=159
a + (n -1)*d = 159
12 + (n-1)*3= 159
(n - 1)*3 = 159 - 12
(n - 1)*3 = 147
n -1 = 147/3
n - 1 = 49
n = 49 + 1
n= 50
Answer: n=50th term
Step-by-step explanation:
An= a1+(n-1)d
An= nth term
A1= first term
N= nth position
D= common difference
An =159
A1= 12
D= 3
N= ?
Substitute the values
An=a1+(n-1)d
159=12+(n-1)3
Collect like terms
159-12=(n-1)3
147=3n-3
147+3=3n
150=3n
N=150/3
N= 50
Therefore 50th term will give 156
B. x+1=0
C. y-1=0
D. y-3=0
I'm answering this assuming that you accidentally wrote 7t instead of 7y.
5x - 8y + 3 = 5x - 7y + 2
- (5x - 8y + 3) -(5x - 8y +3)
0 = y - 1 or y-1 = 0
The answer is C.
Hope I helped!
The product of b and 6 is less than - 16.
Answer:
6b<-16
Step-by-step explanation:
The sentence 'The product of b and 6 is less than -16' translates to the inequality '6b < -16' in Mathematics.
In mathematics, translating a sentence into an inequality involves identifying the mathematical symbols and operations represented by the words in the sentence. In this case, 'the product of b and 6' translates to '6 * b' and 'is less than' translates to '<'. So, the sentence 'The product of b and 6 is less than -16' translates to the inequality '6 * b < -16' or '6b < -16' in simplified form.
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