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1 year ago

Barely learning this in school and not understanding it yet help

Mathematics

2 answers:

omeli [17]1 year ago

8 0

Answer:

The answer is in the picture I hope that is right

Step-by-step explanation:

Lera25 [3.4K]1 year ago

5 0

ANSWER

$x=2184.67$

EXPLANATION

We want to solve the given equation for x:

$-3+8\log _9(3x+7)=29$

First, add 3 to both sides of the equation:

$\begin{gathered} -3+3+8\log _9(3x+7)=29+3 \\ 8\log _9(3x+7)=32 \end{gathered}$

Divide both sides by 8:

$\begin{gathered} \frac{8}{8}\log _9(3x+7)=\frac{32}{8} \\ \log _9(3x+7)=4 \end{gathered}$

Convert from a logarithmic equation to an exponential equation:

$\begin{gathered} \Rightarrow9^4=3x+7 \\ \Rightarrow3x+7=6561 \end{gathered}$

Subtract 7 from both sides of the equation:

$\begin{gathered} 3x+7-7=6561-7 \\ 3x=6554 \end{gathered}$

Divide both sides by 3:

$\begin{gathered} \frac{3x}{3}=\frac{6554}{3} \\ x=2184.67 \end{gathered}$

That is the value of x.

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(3x^4)-5=43 How?? It's supposed to be - 2?

Sergeeva-Olga [200]

Answer:

I got 2 as an answer

Step-by-step explanation:

$( {3x}^{4} ) - 5 = 43 \\ ( {3x}^{4} ) = 43 + 5 \\ ( {3x}^{4} ) = 48 \\ \frac{ {3x}^{4} }{3} = \frac{48}{3}$

${x}^{4} = 16 \\ \sqrt[4]{ {x}^{4} } = \sqrt[4]{16} \\ x = 2$

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}}$

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

savannah is painting a striped background for a mural on a wall that is 15 yards long. she wants a total of 96 stripes that’s ar

adell [148]

In this item, we are tasked to determine the calculation that can be done in order to determine the length or width of the painted mural in each striped area. This can be done by dividing the total length by the number of stripes as shown in the equation below.

Length per stripe = Total length / number of stripes
Length per stripe = 15 yards / 96 stripes
= 0.15625 yard/stripe

8 0

3 years ago

Read 2 more answers

Of a squirrel's hidden nuts, for every 5 that get found, there are 3 that do not get found. A squirrel hid 40 nuts all together.

koban [17]

Answer:

15 nuts do not get found.

Step-by-step explanation:

Given that Of a squirrel's hidden nuts, for every 5 that get found, there are 3 that do not get found.

Total number of nuts squirrel had hidden = 40

Proportion of nuts not found to total = 3/(3+5) = 3/8

Hence out of 40, nuts not found =3/8(40) = 15 nuts.