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3 years ago

(-6,-4) and is parallel to the line x+6y=12

Mathematics

1 answer:

HACTEHA [7]3 years ago

6 0

Answer: y = $-\frac{1}{6}x-5$

Step-by-step explanation:

x+6y=12

6y=-x+12

y=- $\frac{x}{6}$ + 2

If its parallel it means that the slope will be the same

y = mx + b

-4 = -1/6 * (-6) + b

b = -5

Thus,

y = -1/6 x - 5

You might be interested in

Write the percent as a fraction or mixed number in simplest form. 148%

o-na [289]

Answer:

1 and 12/25

its over 100 percent, that's where the 1 comes from. the 48 that's left would be 48/100, which is then simplified to 24/50, which is simplified one more time to 1 12/25

3 0

3 years ago

Read 2 more answers

A checking account rounds the amount of money deposited to the nearest hundredths place. After tax returns on April 15th, Yuan e

ludmilkaskok [199]

Answer:

$\boxed{\text{\$336.35}}$

Step-by-step explanation:

Yuan earned $336.3457.

You are asked to round this number to the nearest hundredth, that is, so there are only two digits after the decimal.

$336.34|57

A rule in rounding numbers is: if the first figure to be dropped is 5 with any nonzero digits following it, then the last figure kept should be rounded up (increased by 1).

The number becomes $336.35.

$\boxed{\large \textbf{\$336.35}} \text{ was placed in Yuan's checking account}$

6 0

4 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}}$

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

3 years ago

Understand multiplying

taurus [48]

Answer:

Step 1 : 1000 * 2 * 9

Step 2 : 1000 * 18

Step 3 : 18000

Step-by-step explanation:

Step 3

2000 * 9 = y ⇒ 18000

y = 18000

Step 1

Solving for this, we have:

Let a be the unknown variable

a * 2 * 9 = 18000

18a = 18000

a = 1000

Therefore, when we multiply 1000 by 2 & 9, we have 18000

Step 2

Solving for this, we have:

Let a be the unknown variable

a * 18 = 18000

18a = 18000

a = 1000

Therefore, when we multiply 1000 by 18, we have 18000

We therefore see that the value of the unknown variables (a, y) is shown above & the product of all the steps (step 1 - 3) is 18000

3 0

4 years ago

(a) A survey of the adults in a town shows that 8% have liver problems. Of these, it is also found that 25% are heavy drinkers,

grandymaker [24]

Answer:

i. Has a liver problems?

= 0.08

ii. Is a heavy drinker ?

= 0.066

iii. If a person is found to be a heavy drinker, what is the probability that this person has liver problem?

= 0.303

iv. If a person is found to have liver problems, what is the probability that this person is a heavy drinker?

= 0.25

v. If a person is found to be a non –drinker, what is the probability that this person has liver problems?

= 0.104

Step-by-step explanation:

We have 2 Events in this question

Event A: People with liver problems

Event B : People without liver problems

Event A: People with liver problems

Let us represent people with liver problems as = (L)

a)8% have liver problems. = P(L)

Under liver problems we have:

b) 25% are heavy drinkers = P( L & H)

c) 35% are social drinkers = P( L & S)

d) 40% are non-drinkers. = P( L & N)

Event B( no liver problem)

Let us represent no liver problem as NL

We are not given in the question but Probability of having no liver problem = 100 - Probability of having liver problem

= 100 - 8% = 92 %

P(NL ) = 92%

From the question, For people without liver problems, we have:

a) 5% are heavy drinkers = P(NL & H)

b) 65% are social drinkers = P( NL & S)

c) 30% do not drink at all = P( NL & N)

An adult is chosen at random, what is the probability that this person

i. Has a liver problems?

P(L) = 8% or 0.08

ii. Is a heavy drinker ?

From the question, we have:

Probability of people that have liver problems and are heavy drinkers P(L & H) = 25% = 0.25

Probability of people that have do not have liver problems and are heavy drinkers P(NL & H) = 5% = 0.05

Probability ( Heavy drinker) =

P(L) × P(L & H) + P(NL) × P(NL & H)

= 0.25 × 0.08 + 0.05 × 0.92

= 0.066

iii. If a person is found to be a heavy drinker, what is the probability that this person has liver problem?

Probability (Heavy drinker and has liver problem) = [P(L) × P(L & H)] ÷ [P(L) × P(L & H)] + [P(NL) × P(NL & H) ]

= [0.25 × 0.08] ÷ [0.25 × 0.08] + [0.05 × 0.92]

= 0.303030303

Approximately = 0.303

iv. If a person is found to have liver problems, what is the probability that this person is a heavy drinker?

P(L & H) = 25% = 0.25

v. If a person is found to be a non –drinker, what is the probability that this person has liver problems.?

People with liver problems are non-drinkers. = P( L & N) = 40% = 0.4

People without liver problems and do not drink at all = P( NL & N) = 30% = 0.3

Probability (non drinker and has liver problem) = [P( L & N) × P(L & H)] ÷ [P( L & N) × P(L & H)] + [ P( NL & N) × P(NL & H) ]

= [0.4× 0.08] ÷ [0.4 × 0.08] + [0.3 × 0.92]

= 0.1038961039

Approximately ≈ 0.104

5 0

3 years ago