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3 years ago

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Petra is asked to color 6/6 of her grid. She must use 3 colors: blue, red ad pink. There must be more blue sections than red sec

tions or pink sections. What are the different ways Petra can color the sections of her grid and follow all the rules?

1 answer:

Phoenix [80]3 years ago

7 0

3 blue 2pink 1red or 3blue 2red 1pink

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WILL MARK BRAINLIEST IF CORRECT!

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A tiger eats 55 pounds of meat a day if he eats 1,754 how many days is that

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897 Days

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3 years ago

If log x+y/5 = 1/2(log x + log y) the find the value of x/y + y/x ?

V125BC [204]

Given that

log (x+y)/5 =( 1/2) {log x+logy}

We know that

log a+ log b = log ab

⇛log (x+y)/5 =( 1/2) log(xy)

We know that log a^m = m log a

⇛log (x+y)/5 = log (xy)^1/2

⇛log (x+y)/5 = log√(xy)

⇛(x+y)/5 = √(xy)

On squaring both sides then

⇛{ (x+y)/5}^2 = {√(xy)}^2

⇛(x+y)^2/5^2 = xy

⇛(x^2+y^2+2xy)/25 = xy

⇛x^2+y^2+2xy = 25xy

⇛x^2+y^2 = 25xy-2xy

⇛x^2+y^2 = 23xy

⇛( x^2+y^2)/xy = 23

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2 years ago

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: <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

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3 years ago

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Sergeeva-Olga [200]

Answer:

x=5 1/4

Step-by-step explanation:

2 1/4= $\frac{9}{4}$ ---> $\frac{9}{4}$ divided by 3 to find 1/7x----> $\frac{3}{4}$ =1/7x---> x=21/4---> x=5 1/4

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