First divide both sides by 5. Then you get the answer. -0 or 0 I'm not sure tho
Answer:
The answer is 3/4
Step-by-step explanation:
When you divide two fractions, such as 1/2 ÷ 2/3, you have to flip the second fraction and then you simply multiply the numerators with each other and the denominator with each other.
Answer:
t = -5
Step-by-step explanation:
Solve for t:
5 (t - 3) - 2 t = -30
Hint: | Distribute 5 over t - 3.
5 (t - 3) = 5 t - 15:
5 t - 15 - 2 t = -30
Hint: | Group like terms in 5 t - 2 t - 15.
Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:
(5 t - 2 t) - 15 = -30
Hint: | Combine like terms in 5 t - 2 t.
5 t - 2 t = 3 t:
3 t - 15 = -30
Hint: | Isolate terms with t to the left hand side.
Add 15 to both sides:
3 t + (15 - 15) = 15 - 30
Hint: | Look for the difference of two identical terms.
15 - 15 = 0:
3 t = 15 - 30
Hint: | Evaluate 15 - 30.
15 - 30 = -15:
3 t = -15
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides of 3 t = -15 by 3:
(3 t)/3 = (-15)/3
Hint: | Any nonzero number divided by itself is one.
3/3 = 1:
t = (-15)/3
Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.
The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:
Answer: t = -5
Answer: 200 bulbs will not be defective.
Step-by-step explanation:
The ratio of defective bulbs to good bulbs produced each day is 2 to 10. This ratio can also be expressed as 1 to 5 by reducing to lowest terms.
The total ratio is the sum of the proportions.
Total ratio = 1 + 5 = 6
This means that if n bulbs is produced, the number of defective bulbs would be
1/6 × n
The number of non defective would be
5/6 × n
Since n = 240, then the number of bulbs that will not be defective is
5/6 × 240 = 200 bulbs
Using a graphing tool
Let's graph each of the cases to determine the solution of the problem
<u>case A)</u>
see the attached figure N 
The range is the interval--------> (0,∞)

therefore
the function
is not the solution
<u>case B)</u> 
see the attached figure N
The range is the interval--------> (0,∞)

therefore
the function
is not the solution
<u>case C)</u>
see the attached figure N
The range is the interval--------> (-∞,3)

therefore
the function
is the solution
<u>case D)</u>
see the attached figure N
The range is the interval--------> (-3,∞)

therefore
the function
is not the solution
<u>the answer is</u>