Help me!! <ABC = ___

Whats the answer two -9=7-x

garri49 [273]

I hope this helps you

7 0

4 years ago

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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

42 =3(2 – 3h) how do you distributive property

ivann1987 [24]

Answer:

h = -4

Step-by-step explanation:

6 0

3 years ago

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6. Solve by factoring: 2x²-11x+14=0

Mariana [72]

Step-by-step explanation:

Solve 2x^2-11x+14=0 give me 2 ways to get the 2 values of x?

The first way is to make factors by middle-term splitting. You get (x-2)(2x-7)=0, so the solution is x=2,7/2.

The other way is to use the formula for the solution of roots. Roots= (-b+ rootD)/2a, (-b-rootD)/2a, if the equation is of the form ax^2+bx+c=0. Here D is the discriminant. D=b^2-4ac

How do I solve 12x^2+11x+2=0?

Solve for x. 5x3−2x2−47x−14=0?

How can I prove that x^2+2x+2=0?

Is the sequence X^2 + 2x -2 =0?

What are the steps to solve 2^(2x)-3(2^x) +2=0?

Lets assume your equation is of the form ax^2+ bx + c =0

First method :

By breaking 'b' factor of x into two parts using factors of 'ac' ( a *c)

like 2* 14 = 28 = 7* 4 ( b is 11 which could be splitted in values 7 and 4)

equation becomes 2x^2 -4x-7x+14 = 0

take out common factors 2x(x-2) -7 (x-2) = 0

which is (2x-7) (x-2) =0 ; x =2,7/2 are two values

Second method:

make equation in the form of (x+h)^ 2 - k = 0 ; then x+h = +sqrt(k) and -sqrt(k)

which will give x as -h+sqrt(k) , +sqrt(k)

2x^2 - 11x +14 =0 wil become x^2 -11/2 x + 7 = 0

which is (x-11/4)^2 + 7- 121/16 =0

(x-11/4)^2 = (121/16) - 7 = 9/16

x- 11/4 = 3/4 and -3/4

x = 14/4 and 8/4 = 7/2 and 2

3 0

2 years ago

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The cost to a store for an MP3 player is $60. The selling price is $1.05. A classmate says that the markup is 175% because $105/

Lapatulllka [165]

MP3 player is $60
The selling price is $105

A classmate says that the markup is 175% because $105/$60=1.75 . Is your classmate correct?

No. IT's not correct

Original price $60
Selling price $105

$105 - $60 = $45

So

45/60 x 100 = 75%

Price is markup 75% from $60 to $105

7 0

4 years ago

Help me!! <ABC = ____?