I really need help Can someone help me

If point A, having

Ronch [10]

gradient or slope=y2 -y1

x2 - x1

so for A, p is x1 and 3 is y1..

For B, 6 is x2 and p is y2

2= p - 3

6 - p

2 ( 6 - p) = p - 3

12 - 2p = p - 3

-2p + p = -3 - 12

-p = - 15

p = 15

4 0

3 years ago

The open intervals on which f is concave upward. (enter your answer using interval notation.

N76 [4]

Khan academy has a lot on this sorry I couldn't answer bro

6 0

3 years ago

Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!

Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}$

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

$\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

$3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x + 3)

$3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$(2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x^2 + 7*x + 12)

$(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)$

$(2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)$

Now we need to solve this:

we will get

$2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)$

$2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9$

Then we get:

$2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0$

$x^3 + 3x^2 + 17*x + 51 = 0$

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

$x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0$

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

6 0

3 years ago

How many duckling were swimming at the pond?

EleoNora [17]

Answer:

There are 20 ducklings.

Step-by-step explanation:

For every group of 6 ducks, 5 would be ducklings.

24 / 6 = 4

There are 4 groups of ducks.

5 * 4 = 20

4 0

3 years ago

True or false: a pair of straight angles can also be adjacent angles

IrinaK [193]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions. the statement "a pair of straight angles can also be adjacent angles" is true

There are some special relationships between "pairs" of angles. Adjacent Angles are two angles that share a common vertex, a common side, and no common interior points. (They share a vertex and side, but do not overlap.) A Linear Pair is two adjacent angleswhose non-common sides form opposite rays.

7 0

3 years ago

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