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3 years ago

I kinda sorta understand this but not completely

Mathematics

1 answer:

Greeley [361]3 years ago

5 0

Problems 1 - 8.
When you multiply the same number raised to exponents, add the exponents. When you divide the same number raised to exponents, subtract the exponents.

1. $7^{-5} \cdot 7^{3} = 7^{-5 + 3} = 7^{-2}$

3. $\dfrac{2^{7}}{2^{10}} = 2^{7 - 10} = 2^{-3}$

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157772+4426891-56 ajutor

arlik [135]

Answer:

4584607 is your correct answer

3 0

3 years ago

What is the difference of the fractions? Use the number line and equivalent fractions to help find the answer.

ipn [44]

The difference of the fraction is $\mathbf{ -\dfrac{3}{4}}$ by using equivalent expressions.

<h3>What is a number line?</h3>

A number line is a diagrammatic representation of real numbers on the graduated line. It ranges from negative axis to positive axis with 0 being the center position.

A sketch showing the representation of the difference in the mixed fractions can be seen in the image attached below.

Using equivalent expressions to determine the difference between the mixed fractions, we have;

$\mathbf{ = -2\frac{1}{2} - (-1\dfrac{3}{4}) }$

$\mathbf{ = -2\frac{1}{2} +1\dfrac{3}{4} }$

$\mathbf{ = -\dfrac{5}{2} +\dfrac{7}{4} }$

$\mathbf{ = -\dfrac{3}{4}}$

Learn more about number line here:

brainly.com/question/25230781

#SPJ1

7 0

2 years ago

Could someone help me and explain?

ExtremeBDS [4]

Answer:

length 3 in
width 2 in

Step-by-step explanation:

Since none of the answer choices match the drawing of the gardener, we assume the question is referring to the drawing of the partner.

The gardener's drawing is 1/4 of actual size. So, in terms of the gardener's drawing, actual size is ...

gardener's drawing = (1/4)actual size

actual size = 4(gardener's drawing)

The partner's drawing is 1/20 of actual size, so is ...

partner's drawing = actual size/20 = (4(gardener's drawing))/20

partner's drawing = (4/20)(gardener's drawing)

partner's drawing = (gardener's drawing)/5

Then the {length, width} of the partner's drawing are ...

partner's drawing {length, width} = {15 in, 10 in}/5 = {3 in, 2 in}

The partner's drawing has a length of 3 inches and a width of 2 inches.

6 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

<u>Differentiation</u> (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

<u>Differential</u> (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

2 years ago

You roll a 6-sided die.

Gnom [1K]

1/6 hope this you :)

8 0

2 years ago