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3 years ago

Helpppppp poooorrrfavooor plsssssss

Mathematics

2 answers:

Trava [24]3 years ago

8 0

Answer:

2/12

Step-by-step explanation:

nikklg [1K]3 years ago

6 0

Answer:

Step-by-step explanation:

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Calculate the perimeter of this shape.<br> 12 cm<br> 18 cm

Lisa [10]

Answer:

$12 +18 + x+y + z +w$

In this special case we know that $x+y =12$ and $z+w =18$ and for this case we can add all the values 12+18 +12+18=60 and that represent the perimeter

Step-by-step explanation:

For this case we know that the perimeter is given by:

$12 +18 + x+y + z +w$

In this special case we know that $x+y =12$ and $z+w =18$ and for this case we can add all the values 12+18 +12+18=60 and that represent the perimeter

6 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

The grocery store sells kumquats for $5.00 a pound and Asian pears for $3.25 a pound. Write an equation in standard form for the

aniked [119]

Okay so your equation is going to be : 18= 5k + 3.25p

3 0

3 years ago

Read 2 more answers

Find the sum of the following ap .1) 1/15, 1/12, 1/10,....,to 11 terms

sertanlavr [38]

Answer:

33/20

Step-by-step explanation:

1/12 - 1/15 = 5/60 - 4/60 = 1/60

d = 1/60

a_n = a_1 + d(n - 1)

a_11 = 1/15 + (1/60)(11 - 1)

a_11 = 1/15 + 1/6

a_11 = 4/60 + 10/60

a_11 = 14/60

a_11 = 7/30

a_12 = 14/60 + 1/60

a_12 = 15/60

a_12 = 1/4

s_n = n(a_1 + a_n)/2

s_11 = 11(1/15 + 7/30)/2

s_11 = 11(2/30 + 7/30)/2

s_11 = 11(9/30)/2

s_11 = 99/60

s_11 = 33/20

8 0

3 years ago

PLEASE HELP ASAP

likoan [24]

Answer:

Step-by-step explanation:

it goes 8 units over and its going up 7

4 0

3 years ago