The expression is equivalent to the given expression is: negative 9 over 12 times t minus 2 over 9.
<h3>What is stated as the
equivalent expression?</h3>
- Equivalent expressions have been expressions that perform the same function despite their appearance.
- If two linear algebra expressions are equivalent, they have the same value when the variable is set to the same value.
For the given question, the expression is stated as-
Negative 7 over 6 times t plus 4 over 9 end quantity minus expression quantity negative 5 over 12 times t plus 2 over 3 end quantity
This can be written as;
= -7t/6 + 4/9 - (-5t/ 12 + 2/3)
Simplify the expression as;
= -7t/6 + 4/9 + 5t/ 12 - 2/3
Re arrange the numbers
= -7t/6 + 5t/ 12 - 2/3 + 4/9
Combine the variables and constant separately.
= -9t/12 - 2/9
This can be written in words as- negative 9 over 12 times t minus 2 over 9.
Thus, the expression is equivalent to the given expression is: negative 9 over 12 times t minus 2 over 9.
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Answer:
C = 13
Step-by-step explanation: According to the pythagorean theorem, which states A² + B² = C². This means that in this equation 5²+12²=x². When we solve this we get 169=x². Since 169 is the perfect square of 13 therefore the answer is 13.
Answer:
1,320 seconds or 22 minutes to walk 1 mile
Step-by-step explanation:
5280 feet (1 mile) ÷ 4 feet per second = 1,320 seconds or 22 minutes
Answer:
rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour
Step-by-step explanation:
We will make a table of the trip there and back using the formula distance = rate x time
d = r x t
there
back
The distance there and back is 264 miles, so we can split that in half and put each half under d:
d = r x t
there 132
back 132
It tells us that the trip there is with the wind and the trip back is against the wind:
d = r x t
there 132 = (r + w)
back 132 = (r - w)
Finally, the trip there took 3 hours and the trip back took 6:
d = r * t
there 132 = (r + w) * 3
back 132 = (r - w) * 6
There's the table. Using the distance formula we have 2 equations that result from that info:
132 = 3(r + w) and 132 = 6(r - w)
We are looking to solve for both r and w. We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:
132 = 3r + 3w and
132 - 3w = 3r so
44 - w = r. Subbing that into the second equation:
132 = 6(44 - w) - 6w and
132 = 264 - 6w - 6w and
-132 = -12w so
w = 11
Subbing w in to solve for r:
132 = 3r + 3(11) and
132 = 3r + 33 so
99 = 3r and
r = 33
