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2 years ago

Help please.. z zzz z Z z z z z z z z z z z z z zz

Mathematics

2 answers:

Arisa [49]2 years ago

8 0

The answer is option B.

Rudiy272 years ago

6 0

It’s B I believe …..

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Use the distributive property to simplify this expression 1/2(4x+2) + 2x +3

mel-nik [20]

Answer:

4x + 4

Step-by-step explanation:

8 0

2 years ago

Helpppppp i give you 98 points !!!!!!!!!!!!Using the distributive property to find the product (y−4x)(y2+4y+16) results in a pol

Marina CMI [18]

Answer:

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor, that is

y(y² + 4y + 16) - 4x(y² + 4y + 16) ← distribute both parenthesis

= y³ + 4y² + 16y - 4xy² - 16xy - 64x

Compare - axy to - 16xy ⇒ a = 16

4 0

3 years ago

Read 2 more answers

99 Points for correct answer!!!!!!!!!

3241004551 [841]

The equation given is:

15x - 3y = 21

Slope intercept form is:

y = mx + b

15x - 3y = 21

-15x -15x

-------------------

-3y = -15x + 21

/-3 /-3 /-3

---------------------

y = 5x -7

The answer is: y = 5x - 7

Hope I helped!

-Char

3 0

3 years ago

A tap takes 3 hours to fill a tank. A discharge tap takes 6 hours to empty the same tank. How long will it take to fill the tank

wlad13 [49]

Answer:

2 hours

Step-by-step explanation:

first tap= 3hours

therefore work done is = 1/3

second tap= 6hours

therefore work done = 1/6

so, 1/3+ 1/6

=2/6 + 1/6

=3/6

=1/2

=2/1 hours

=2 hours

8 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

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

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

3 0

3 years ago