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3 years ago

7

How many solutions are there to the equation, 220=2x-12x

2 answers:

galben [10]3 years ago

8 0

Hi,

For the equation: 220=2x-12x there is only one solution, x=−22.

Have a great day!

PolarNik [594]3 years ago

8 0

Answer: 1

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

9. An article was sold at a profit of 12%. Had it been sold for Nu 80 more, there would be a profit of

Kamila [148]

Answer:

<em>The cost price of the article was Nu 1,000</em>

<em>The original selling price was Nu 1,120</em>

<em>The second selling price was Nu 1,200</em>

Step-by-step explanation:

<u>System of Equations</u>

Let's call:

x = Original selling price of the article

y = Cost of the article

The first condition implies that the difference between the selling price and the cost price (the profit) is 12%. Recall the profit % is always calculated with respect to the cost, that is, the profit at the first condition is:

0.12y = Original profit

Setting up the equation, we have:

$x-y=0.12y$

The second condition states the selling price is 80 Nu more than before, that is:

x+80= Second selling price of the article

y = Cost of the article (it doesn't change)

The profit of this condition is 20%, thus:

0.2y = Second profit

This produces a second equation:

$x+80-y=0.2y$

Let's put both equations together to form a system of equations:

$x-y=0.12y$

$x+80-y=0.2y$

Operating on both equations:

$x=1.12y$

$x=1.2y-80$

Equating both equations, we have an equation with only y's:

$1.12y=1.2y-80$

Subtracting 1.2y on each side:

$1.12y-1.2y=-80$

Operating:

$-0.08y=-80$

Solving:

$\displaystyle y=\frac{-80}{-0.08}=1,000$

a)

The cost price of the article was Nu 1,000

b) The selling price of the article will be calculated in both cases:

Original selling price: x=1.12y= 1.12*1000=1,120

The original selling price was Nu 1,120

The second selling price was x+80=1,200

The second selling price was Nu 1,200

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3 years ago

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vitfil [10]

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3 years ago

How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

Zarrin [17]

By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

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3 years ago