Write an equation, solve and show all work.

Solve for x. A. x = 12 B. x = 4 C. x = 3 D. x = 1

kakasveta [241]

Answer:

4=x

Step-by-step explanation:

1/5( 2x+92) = 5x

Multiply each side by 5

5*1/5( 2x+92) = 5x*5

( 2x+92) = 25x

Subtract 2x from each side

2x+92-2x = 25x-2x

92 = 23x

Divide by 23

92/23 = 23x/23

4=x

7 0

3 years ago

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mackenzie made 33 out of 50 throws. what is the experimental probability that mackenzie will not make the next free throw she at

siniylev [52]

answer:

1.94117647058

Step-by-step explanation:

33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058

5 0

3 years ago

use the quadratic formula to slove the equation. if necessary, round to the nearest hundredth. x^2+3=-4x

nexus9112 [7]

Answer: The zeros are x= -1 and x= -3

7 0

3 years ago

Match each linear equation to the appropriate description: Column A 1. : 2. : 3. : 4. : Column B a. A linear equation with slope

timama [110]

b. A linear equation with slope 5 and y-intercept 1 c. A linear equation with slope 2 and y-intercept 3 d. A linear equation with slope 3 and y-intercept 2

4 0

3 years ago

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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