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3 years ago

25% discount off of $350 Original: Discount: New price: Plz help

Mathematics

2 answers:

Eduardwww [97]3 years ago

6 0

I think original price is $350

the discount will be 25% of 350 so it will be 25/100 x 350 to get 87.5 dollars

the new price's percentage is 100% -25% so to get the new price it'll be 75% of 350 to get 262.5 dollars

devlian [24]3 years ago

4 0

Original:350 discount:87.50 off new price: 262.5

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Natasha2012 [34]

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3 years ago

Factor using the x method ( please do not answer without showing work )

muminat

Answer:

$5(x + 10)(10x - 3)$

Step-by-step explanation:

We are factoring

$50x^{2} + 485x - 150$

So:

((2•5^2x^2) + 485x) - 150

Pull like factors :

50x^2 + 485x - 150 = 5 • (10x^2 + 97x - 30)

Factor

10x^2 + 97x - 30

Step-1: Multiply the coefficient of the first term by the constant 10 • -30 = -300

Step-2: Find two factors of -300 whose sum equals the coefficient of the middle term, which is 97.

-300 + 1 = -299

-150 + 2 = -148

-100 + 3 = -97

-75 + 4 = -71

-60 + 5 = -55

-50 + 6 = -44

-30 + 10 = -20

-25 + 12 = -13

-20 + 15 = -5

-15 + 20 = 5

-12 + 25 = 13

-10 + 30 = 20

-6 + 50 = 44

-5 + 60 = 55

-4 + 75 = 71

-3 + 100 = 97

Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 100

10x^2 - 3x + 100x - 30

Step-4: Add up the first 2 terms, pulling out like factors:

x • (10x-3)

Add up the last 2 terms, pulling out common factors:

10 • (10x-3)

Step-5: Add up the four terms of step 4:

(x+10) • (10x-3)

Which is the desired factorization

Thus your answer is

$5(x + 10)(10x - 3)$

8 0

3 years ago

Read 2 more answers

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MakcuM [25]

75.09 = seventy- five and nine-hundredths

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3 years ago

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Doreen is flipping two fair coins. What is the probability that both coins land on heads?

andrezito [222]

Answer:

There are four different possible outcomes: both coins are heads, the red coin is heads and the blue coin is tails, the red coin is tails and the blue coin is heads, or both coins are tails. Each outcome has equal probability. So the probability of both being heads is 1/4.

Step-by-step explanation:

6 0

3 years ago

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.

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4 years ago