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3 years ago

Jorge wants to buy an 8-pack of toothbrushes. The price for the toothbrushes is

Mathematics

2 answers:

fgiga [73]3 years ago

7 0

Answer: $1.89

Step-by-step explanation: I did the math and got this :/

Scrat [10]3 years ago

3 0

Answer:

$18.12 - $3.00 = 15.12 / 8 = 1.89 per toothbrush

Step-by-step explanation:

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Lisa is cooking muffins. The recipe calls for 7 cups of sugar. She has already put in 2 cups. How many more cups does she need

Tresset [83]

the answer is 5. 7-2 is 5.

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3 years ago

Triangle XYZ is a right triangle. if XY=4, what is YZ

ruslelena [56]

Well what is XZ. If XY is 4 & YZ is undefined. Is YZ undefined also? or does it have a given measurement?

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2 years ago

Read 2 more answers

Convert the following polar-form vectors of the form ⟨r,θ⟩ into component-form vectors of the form ⟨x,y⟩. Your answers should be

storchak [24]

9514 1404 393

Answer:

a) <8.356, 9.959>

b) <-0.605, -1.663>

c) <-5.023, 2.9>

Step-by-step explanation:

Many calculators can perform polar ⇔ rectangular conversion. Attached is the result from one of them. Of course, you can also program a spreadsheet to do it. (The ATAN2( ) function is useful for finding the correct angle.)

If you want to do these calculations by hand, the conversion is ...

<r, θ> ⇒ <r·cos(θ), r·sin(θ)>

In the attached, the rectangular coordinates are shown as complex numbers. The imaginary component is the y-component of the vector.

8 0

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>

3 0

3 years ago

Help me fast asap this is due by the end of class

Rufina [12.5K]

Answer:

Step-by-step explanation:

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x^10/x^8= x^2

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5 0

1 year ago