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Gre4nikov [31]
3 years ago
5

A number, y, is equal to the difference of a larger number and 3. The same number is one-third of the sum of the larger number a

nd 9. Which equations represent the situation? A. x + y = 3 and x minus 3 y = negative 27 B. x + y = 3 and x minus 3 y = negative 9 C. x minus y = 3 and x minus 3 y = negative 27 D. x minus y = 3 and x minus 3 y = negative 9
Mathematics
2 answers:
Shalnov [3]3 years ago
4 0

Answer:

D. x - y = 3 and x - 3 y = - 9

Step-by-step explanation:

The descriptions above describe the equation of choice D, so I pick this answer.

professor190 [17]3 years ago
3 0

Answer:

D

Step-by-step explanation:

just took the test

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Using the 45°-45°-90° triangle theorem, find the value of h, the height of the wall. 6. 5 ft 6. 5 StartRoot 2 EndRoot ft 13 ft 1
navik [9.2K]
35 es mucho así que será 27 i think is 18
4 0
2 years ago
Help Please This Is a bit hard so I will give Brainliest if Correct!!
Goshia [24]

Answer:

125 to the 6th power.

Step-by-step explanation:

2 * 3 * 125 = 750

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7 0
3 years ago
A CD has a 5% chance of being a smash hit and profiting $5.2 million, a 50% chance of being a modest success and profiting $0.9
Zinaida [17]
The probability of profiting $5.2 million is 5% = 0.05

The probability of profiting $0.9 million is 50% = 0.5

The probability of breaking even, meaning no profit gain, is 45% = 0.45

Expected profit values of the CD is given by
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3 0
3 years ago
1.<br> Find the product of -12 and -6.
Alborosie

Answer:

72

Step-by-step explanation:

Multiply. Note that when you multiply two negative numbers, the answer will be positive.

Combine:

-12 * -6 = 12 * 6 = 72

72 is your answer.

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5 0
3 years ago
Lim t^4 - 6 / 2t^2 - 3t + 7
Harman [31]

I think you meant to say

\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}

(as opposed to <em>x</em> approaching 2)

Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}

Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}

6 0
3 years ago
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