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3 years ago

Help and i will make brainliest and will than and rate and give points if someone helps me

Mathematics

1 answer:

Marysya12 [62]3 years ago

6 0

1.
$8{c}^{2} - 1.5d \\$
2.
$6 - 4x$
3. -7/30
4. i believe it is the communitive of multiplication , but not 100% sure

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What is the product when 20 3 is multiplied with 9 25 ?

Novay_Z [31]

Answer:

187.775

Step-by-step explanation:

20.3×9.25=

187.775

4 0

3 years ago

64. Denise wants to burn at least 5000 calories a week through running. Based on her running speed, she estimates that she can b

Rudiy27

An inequality that represents Denise's goal in terms of the number of hours spent running 'h'

$550h\leq 5000$

Given :

Denise wants to burn at least 5000 calories a week through running.

she can burn 550 calories per hour

Let 'h' be the number of hours spent

In 1 hour she can burn 550 calories

In 'h' hours she can burn 550h calories

Given that she want to burn atleast 5000 calories in a week

Burn atleast 5000 calories means <=5000 calories

So , the inequality that represents Denise's goal is

calories burn in h hours <= 5000 calories

$550h\leq 5000$

Learn more : brainly.com/question/381815

5 0

2 years ago

(5x+3)(7x-7) how do find x

Delvig [45]

We can use the FOIL method to solve.

(5x + 3)(7x - 7)

(5x * 7x) + (5x * -7) + (3 * 7x) + (3 * -7)

35x - 35x + 21x - 21

0 + 0

Best of Luck!

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 too easy 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.

Differential (Power Rules)

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

Differentiation (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)}}$

Differential (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}}$