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azamat
3 years ago
14

4y−7+2y=3(y−1)−1 answer for y

Mathematics
2 answers:
Paladinen [302]3 years ago
4 0

Answer:

y = 1

Step-by-step explanation:

4y−7+2y=3(y−1)−1

6y −7 = 3y - 3 - 1

6y −7 = 3y - 4

6y = 3y + 3

3y = 3

y = 1

sammy [17]3 years ago
3 0
The answer to your question is y=1
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93% of what is 198 tons?<br><br> I dont get this someone help
Vera_Pavlovna [14]
93% of x = 198 tons
93% = 93/100
= 0.93
0.93 * x = 198 tons

Divide both sides by 0.93:
x = 198/0.93
x = 212.90 -> Final Answer.
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3 years ago
If b is a positive real number and m and n are positive integers, then
jenyasd209 [6]

Answer:

true

Step-by-step explanation:

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3 years ago
The temperature is at -2 F at 6A.M. During the next three hours, the temperature increases to 15F. What is the temperature at 9A
egoroff_w [7]

Answer:

Temperature at 9 A.M =  13° F

Step-by-step explanation:

Given:

Temperature at 6 A.M = - 2° F

Temperature increases in 3 hours =  15° F

Find:

Temperature at 9 A.M

Computation:

Temperature at 9 A.M = Temperature at 6 A.M + Temperature increases in 3 hours

Temperature at 9 A.M =  - 2° F + 15° F

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6 0
3 years ago
Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=
jasenka [17]

Answer:

The function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

The function is continuous at point x=36 if:

  1. The function \\ f(x) exists at x=36.
  2. The limit on both sides of 36 exists.
  3. The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}

\\ f(36) = \frac{36*6}{900} = \frac{6}{25}

The limit of the \\ f(x) is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Since

\\ f(36) = \frac{6}{25}

And

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Then, the function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

8 0
3 years ago
What is the least common denominator of 5/6 and 3/8?​
VARVARA [1.3K]

Answer:

Step-by-step explanation:

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8 0
3 years ago
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