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3 years ago

The rule T 5, -0.5° Ro, 1800(x, ) is applied to FGH to

Mathematics

1 answer:

Yuki888 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

Given rule for the multiple translations is,

$T_{5,-0.5}.R_{0.180^{\circ}}(x,y)$

Apply the rule $R_{0,180^{\circ}}$ first.

(x, y) → (-x, -y)

This rule illustrates a rotation of the triangle FGH by 180° about the origin,

Vertices of ΔFGH are,

F → (1, 1)

G → (4, 5)

H → (5, 1)

After rotation vertices of the image triangle are,

F' → (-1, -1)

G' → (-4, -5)

H' → (-5, -1)

Further apply the rule,

$T_{5,-0.5}$

(x, y) → (x + 5, y - 0.5)

By this rule of translation,

F'(-1, -1) → F"{(-1 + 5), (-1 - 0.5)}

→ F"(4, -1.5)

G'(-4, -5) → G"[(-4 + 5), (-5 - 0.5)]

→ G"(1, -5.5)

H'(-5, -1) → H"[(-5 + 5), (-1 -0.5)]

→ H"(0, -1.5)

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Find the derivative of ln(secx+tanx)

Sliva [168]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000160

————————

Find the derivative of

$\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}$

You can treat y as a composite function of x:

$\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.$

so use the chain rule to differentiate y:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}$

The first derivative is 1/u, and the second one can be evaluated by applying the quotient rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}$

Multiply out those terms in parentheses:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Substitute back for $\mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Simplifying that product, you get

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}$

∴ $\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus

3 0

3 years ago

The diameter of Saturn at its equator is about 120,540,000 meters. What is 120,540,000 written in word form?

Marianna [84]

Answer:

Step-by-step explanation:

Onehundred twenty-million-fivehundred forty thousand

8 0

3 years ago

Express 18 as a fraction of 27. Give your answer in its simplest form

den301095 [7]

Answer:

Thus, 2/3 is the simplified fraction for 18/27 by using the GCD or HCF method. Thus, 2/3 is the simplified fraction for 18/27 by using the prime factorization method

Step-by-step explanation:

hope l am not late!

6 0

3 years ago

Dr Wellbee advised his patients to drink 8 eight-ounce glasses of water everyday for every 50 pounds of weight. Kelby drank 168

BlackZzzverrR [31]

Kelby drank 168 glasses of water per week
i.e 24 glasses per day

Since 8 eight-ounce glass of water is for every 50 pounds.
1 eight-ounce glass is for = 50/8 pounds
and 24 glasses of water is for = 50*24/8
= 150pounds

So kelby probably weighs 150 pounds

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3 years ago

Every _____ tessellates. A. Decagon B. Polygon C. Triangle D. Hexagon

olga2289 [7]

The answer is letter B. Every polygon tesselates. Since the word 'tesselate' here means a pattern of shapes put together without any gaps, then I think every polygon tesselates if they are put side by side together in varying amounts and leave absolutely no gaps or spaces between each other.

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3 years ago