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3 years ago

Two ways to solve 2(4x-11)=10

1 answer:

3 0

Hi there,

2 (4x - 11) = 10
4x - 11 = 5
4x = 5 + 11
4x = 16 then divide both sides by 4
16 divided by 4 = 4

Hope this helps :)

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Find X. Round to the nearest tenth. Law of Cosines : c2 = 22 + b2 - 2ab cos C

fiasKO [112]

Answer:

70.5°

Step-by-step explanation:

22² = (20)²+(18)² - 2(20)(18) cos X

484 = 400 + 324 - 720 cos X

-240 = -720 cosx

1/3 = cos X

$cos^{-1}(\frac{1}{3})$ = X

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7 0

3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

Hence the limit of the function $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

Apply the distributive property to 9(x + 2)

Anni [7]

Uh it would be 9x+18.

8 0

3 years ago

Read 2 more answers

If you were applying for a job and the boss said: “We have 7 employees and the average salary is $100,000,” would you hope that

frosja888 [35]

Mode, because mode is the number that shows up most often.

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3 years ago

Which phrase describes the variable expression z+ 8?

a_sh-v [17]

9514 1404 393

Answer:

A. z increased by 8

Step-by-step explanation:

When a positive value is added, the original value (z) is increased.

z + 8 ⇔ z increased by 8

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3 years ago