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ss7ja [257]
2 years ago
8

Find the equation of the line that goes through the point (0, 4) and is parallel to the line y=3x

Mathematics
1 answer:
den301095 [7]2 years ago
7 0

\huge\boxed{\sf{Hi\;there!}}

Parallel lines have the same slope.

The given line has a slope of 3, therefore, the line parallel to the given line has a slope of 3.

Now, we have the slope of the line and a point that it passes through.

So, we can use the Point-Slope formula:

\bigstar \sf{y-y1=m(x-x1

\bigstar \sf{y-4=3(x-0)

\bigstar \sf {y-4=3x

\bigstar \sf {y=3x+4

\huge\boxed{\boxed{Answer:\;y=3x+4}}}

\huge\bold{Hope\;it\;helps\;you.\;Good\;luck\;and\;have\;a\;nice\:day.}

\boxed{AmiableTeen:)}

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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
Tanya's car will go 45 meters on 7 gallons. Tanya wants to know how fuel efficient the car is. Please help by computing the rati
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Answer:

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fuel efficiency of Car is 7 meters per gallons

Step-by-step explanation:

Given Tanya car goes 45 meters on 7 gallons.

In terms of ratio  of distance traveled and fuel consumed

45 meters : 7 gallons

since both 45 and 7 are multiple of 7 dividing both side by 7

45/7 meters : 7/7 gallons

5 meters : 1 gallon

Thus, fuel efficiency of Car is 7 meters per gallons.

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Write an inequality to match the problem.<br> There are at least 28 days in every month.
Elodia [21]

Answer:

28 ≥ month

Step-by-step explanation:

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