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3 years ago

G(x)=2x^2+3x+6 find g(p+3)

1 answer:

5 0

We're finding the function of (p+3), which means wherever there's an x, we're replacing it with (p+3).
2 (p+3)^2 + 3 (p+3) + 6
Foil out (p+3)^2.
2 (p^2 + 6p + 9) + 3 (p+3) + 6
Distribute.
2p^2 + 12p + 18 + 3p + 9 + 6
Add like terms.
g (p+3) = 2p^2 + 15p + 33

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Identify the property used in the first five steps to solve the equation 2(x – 5) - 6x= -22

USPshnik [31]

To answer this question, you need to multiply the number inside the bracket first. Then you can move the number to the right side of the equal sign and keep the x on the left side of the equal sign. The step would be like this

2(x – 5) - 6x= -22
(2x - 10) - 6x = -22
2x - 6x = -22 +10
-4x= -12
x= -12/-4
x=3

4 0

3 years ago

I need help fast. please help me

galina1969 [7]

ANSWER :

I THINK ITS B

EXPLANATION :

ANSWERED BY THE BEST !!!!

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7 0

3 years ago

Read 2 more answers

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by reductio ad absurdum that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

3 years ago

Two sides of a triangle measure 25cm and 35cm. which of the following could be the measure of the 3rd side

Bumek [7]

11 cm? Im not that good at this.. If im correct any of those would work

8 0

3 years ago

A farmer wants to fence a rectangular habitat for goats. The length (y) of the habitat has to be at least 80 feet. The farmer cu

Leni [432]

Answer:

d. The farmer can make a habitat that is 40 feet wide and 100 feet long.

Step-by-step explanation:

The attached graph shows the inequalities and the various answer choices. The existence of the doubly-shaded area shows that answer choice C is false.

The appropriate choice is ...

d. The farmer can make a habitat that is 40 feet wide and 100 feet long.

4 0

3 years ago