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3 years ago

Y = sec θ tan θ differentiate

1 answer:

6 0

Y = sec t tan t = 1/ cos t * sin t / cos t = sin t / cos² t
$y`= \frac{cos t cos ^{2} t-2 cost*(-sin t)*sint}{cos^{4} t}= \\ =\frac{cos ^{3} t+2costsin^{2}t }{cos ^{4}t }= \\ = \frac{cos ^{2} t+2sin ^{2} t}{cos ^{3} t}= \\ = \frac{1-sin ^{2}t+2sin ^{2} t }{cos ^{3}t }= \\ = \frac{1+sin ^{2} t}{cos ^{3}t }$

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The length of a rectangle is 2 ft longer than its width.

tino4ka555 [31]

Hi! I'm happy to help!

To solve this, we first need to look at the perimeter equation:

P=2L+2W

We don't know our length, so we can represent it with x. Since our width is 2 feet shorter than x, we can represent it with x-2. Now, we plug these values into our equation:

56=2x+(2(x-2))

Let's simplify what the width is by multiplying:

56=2x+2x-4

Now, let's combine our 2xs

56=4x-4

Now, we just need to solve for x in order to find our length and width.

First, we need to isolate x on one side of the equation. We can do this by adding 4 to both sides:

56=4x-4

+4 +4

60=4x

Now, all we have to do is divide both sides by 4 and x will be fully isolated:

60=4x

÷4 ÷4

15=x

Now that we know x, let's plug this into our previous equations:

L=x=15

W=x-2=15-2=13

To verify our answers, we can plug this into our perimeter equation:

56=2(15)+2(13)

56=30+36

56=56

After double checking our answers, we know that our length is 15 and our width is 13.

I hope this was helpful, keep learning! :D

5 0

2 years ago

Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui

Vesnalui [34]

<h3>The dimensions of the given rectangular box are:</h3><h3>L = 15.874 cm , B = 15.874 cm , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box

= Area of the base + 4 x ( Area of the adjacent faces)

= S x S + 4 ( S x H) = S² + 4 SH ..... (1)

Also, Area of the box = S x S x H = S²H

⇒ S²H = 2000

$\implies H = \frac{2000}{S^2}$

Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S = 15.874 cm in the value of H , we get:

$\implies H = \frac{2000}{S^2} = \frac{2000}{(15.874)^2} = 7.8937 cm$

Hence, the dimensions of the given rectangular box are:

L = 15.874 cm

B = 15.874 cm

H = 7.8937 cm

3 0

3 years ago

Given that f ( x ) = 8 x + 5 f ( x ) = 8 x + 5 and g ( x ) = 3 − x 2 g ( x ) = 3 - x 2 , calculate f(g(0))

Alex17521 [72]

Answer:

Step-by-step explanation:

g(0)=3-2(0)

= 3

f(3) = 8(3)+5

f(3) = 29

5 0

2 years ago

Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.

Mkey [24]

Answer:

$1.~e^{-2} \\2.~e^{\frac{7}{2}}\\3.~e^8\\4.~e^{\frac{-11}{2}}$

Step-by-step explanation:

We have to simplify the given exponential exponents.

Exponential Properties:

$e^0 =1\\e^a.e^b = e^{a+b}\\\\\displaystyle\frac{e^a}{e^b} = e^{a-b}\\\\(e^a)^b = e^{ab}\\\\e^{-a} = \frac{1}{e^a}$

Simplification takes place in the following manner:

$(e^{-3})^\frac{2}{3}\\(e^a)^b = e^{ab}\\=e^{-3\times \frac{2}{3}}\\=e^{-2}$

$(e^4)(e^{\frac{-1}{2}})\\e^a.e^b = e^{a+b}\\=e^{(4+\frac{-1}{2})} \\= e^{\frac{7}{2}}$

$(e^{-2})^{-4}\\(e^a)^b = e^{ab}\\= e^{-2\times -4}\\=e^8$

$(e^{-4})(e^{\frac{-3}{2}})\\e^a.e^b = e^{a+b}\\=e^{(-4+\frac{-3}{2})} \\= e^{\frac{-11}{2}}$

4 0

3 years ago

Having trouble with my functions homework plz can someone help

puteri [66]

Answer:

$g(x) = \frac{ - 19}{3} x \: + 2$

I hope that is right

4 0

3 years ago