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3 years ago

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This is a cross-sectional view of candy bar ABC. A candy company wants to create a cylindrical container for candy bar ABC so th

at it is circumscribed about the candy bar. If segment AD = 2.5 cm, what is the smallest diameter of wrapper that will fit the candy bar? ABC in which point E is between points A and B, point D is between points A and C, point F is between points B and C, segments AE and EB are congruent, segments BF and FC are congruent, and angles AED, ABF, and DFC are right angles. A. 3 cm B. 4cm C.5cm D.6cm

1 answer:

Komok [63]3 years ago

6 0

Answer:

a3

Step-by-step explanation:

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Need help fast plz for my pre calc exam

bezimeni [28]

Answer:

<em>h = 8.54 units</em>

Step-by-step explanation:

<u>The Law of Cosines </u>

It relates the length of the sides of a triangle with one of its internal angles.

Let a,b, and c be the length of the sides of a given triangle, and x the included angle between sides a and b, then the following relation applies:

$c^2=a^2+b^2-2ab\cos x$

Since we know the values of all three side lengths, we solve the equation for x:

$\displaystyle \cos x=\frac{a^2+b^2-c^2}{2ab}$

For the triangle ABC in the image, a=10, b=15, c=13, thus:

$\displaystyle \cos A=\frac{10^2+15^2-13^2}{2*10*15}$

$\displaystyle \cos A=\frac{156}{300}$

$\displaystyle \cos A=0.52$

Thus, A = 58.67°

For the right triangle of height h and hypotenuse 10, we use the sine ratio:

$\displaystyle \sin 58.67^\circ=\frac{h}{10}$

Solving for h:

$h = 10\sin 58.67^\circ$

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3 years ago

Solve for b.<br> 39 = 28 - 6

prohojiy [21]

Answer:

39=22

28-6 equal to 22 so 39= 22

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3 years ago

Read 2 more answers

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

tatyana61 [14]

Answer:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

Step-by-step explanation:

So we have:

$y\sin(y)=x\cos(x)$

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

$\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[y\sin(y)]$

We can use the product rule:

$(uv)'=u'v+uv'$

So, our derivative is:

$=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]$

We must implicitly differentiate for y. This gives us:

$=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]$

For the sin(y), we need to use the chain rule:

$u(v(x))'=u'(v(x))\cdot v'(x)$

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

$=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})$

Simplify:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}$

And we are done for the right.

Right Side:

We have:

$\frac{d}{dx}[x\cos(x)]$

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

$=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]$

Differentiate:

$=\cos(x)-x\sin(x)$

So, our entire equation is:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)$

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

$\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)$

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

And we're done!

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3 years ago

What is the area of this trapezoid?

kolezko [41]

Answer:

are of trapezoid =1/2×height(a+b)

Step-by-step explanation:

A=1/2×8(10+20)

=4×30

=120cm*

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3 years ago

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Pachacha [2.7K]

Answer:

A

Step-by-step explanation:

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Given

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The term - $\frac{1}{x}$ means that f(x) is not a polynomial

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3 years ago