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My name is Ann [436]
3 years ago
5

Please hepl with math

Mathematics
1 answer:
Ivenika [448]3 years ago
6 0

Answer:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

Step-by-step explanation:

We are given the expression to be simplified:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}}

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.

We know that:

a^{p+q} = a^p \times a^q

Let us use it to solve the powers of 5 in the given expression.

\therefore we can write:

5^{n+2} = 5^{n+1}\times 5= 5^n\times 5^{2}

5^{n+1} = 5^n\times 5

The given expression becomes:

\dfrac{5^{n+1} \times 5-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n}\times 5}

Taking common 5^{n+1} from the numerator and

Taking common 5^{n} from the denominator

\Rightarrow \dfrac{5^{n+1} (5-6)} {5^{n}(13-2\times5)}\\\Rightarrow \dfrac{5^{n+1} (-1)} {5^{n}(13-10)}\\\Rightarrow -\dfrac{5^{n+1}} {5^{n}\times3}\\\Rightarrow -\dfrac{5^{n}\times 5} {5^{n}\times3}\\\Rightarrow -\dfrac{5}{3}

\therefore The answer is:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

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What is the equation of the line that passes through the points (5, 3) and (-3,-1)?
Liono4ka [1.6K]

Answer:

y=1/2x+1/2

m=1/2

Step-by-step explanation:

You want to find the equation for a line that passes through the two points:

(5,3) and (-3,-1).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (5,3), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=5 and y1=3.

Also, let's call the second point you gave, (-3,-1), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=-3 and y2=-1.

Now, just plug the numbers into the formula for m above, like this:

m=

-1 - 3 over

-3 - 5

or...

m=

-4 over

-8

or...

m=1/2

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=1/2x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(5,3). When x of the line is 5, y of the line must be 3.

(-3,-1). When x of the line is -3, y of the line must be -1.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=1/2x+b. b is what we want, the 1/2 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (5,3) and (-3,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(5,3). y=mx+b or 3=1/2 × 5+b, or solving for b: b=3-(1/2)(5). b=1/2.

(-3,-1). y=mx+b or -1=1/2 × -3+b, or solving for b: b=-1-(1/2)(-3). b=1/2.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points

(5,3) and (-3,-1)

is

y=1/2x+1/2

7 0
3 years ago
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