PLEASE HELP MEEE Write a linear function that related y to x

Rectangle PQRS will be transformed to the points P^ prime (2,0) , Q^ prime (7,0),R^ prime (7,-2) , and S^ prime (2,-2)What type

Svet_ta [14]

I think it’s D rotation can u plz thank me

6 0

2 years ago

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Cmon fwm on dis one to

stich3 [128]

Answer:

x = 10.5

Step-by-step explanation:

we know that ZP = PX (its the midpoint of the rectangle)

11 = PX

ZP + PX = ZX

11+11 = ZX

22 = ZK

2x + 1 = 22

2x = 22 -1

2x = 21

x = 21/2

x = 10.5

3 0

2 years ago

The sum of an infinite geometric series is three times the first term. Find the common ratio of this series.

adelina 88 [10]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

The sum of an infinite geometric series is expressed according to the formula;

$S_\infty = \dfrac{a}{1-r}$ where;

a is the first term of the series

r is the common ratio

If the sum of an infinite geometric series is three times the first term, this is expressed as $S_\infty = 3a$

Substitute $S_\infty = 3a$ into the formula above to get the common ratio r;

$3a = \dfrac{a}{1-r} \\\\$

$cross \ multiply\\\\3a(1-r) = a\\\\3(1-r) = 1\\$

open the parenthesis

$3 - 3r = 1\\\\$

subtract 3 from both sides

$3 - 3r -3= 1-3\\\\-3r = -2\\\\r = \frac{2}{3}$

Hence the common ratio of this series is $\frac{2}{3}$

3 0

3 years ago

Can someone tell me what 2/3 divided by 5 is it’s for a quiz.

docker41 [41]

Answer:2/15 hope this helps

Step-by-step explanation:

7 0

3 years ago

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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

2 years ago