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3 years ago

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You are ordering 100 muffins. The buisness allows you to choose 5 different kinds out of their 12 kinds. How many ways can you c

hoose 5 types of muffins from a variety of 12 types

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

12 divide 5

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Which similarity statements describe the relationship between the two triangles? Check all that apply.

astra-53 [7]

Answer:

ABC is similar to DEF

ABC is similar to DFE

Step-by-step explanation:

6 0

3 years ago

What are the domain and range of f (x) = log (x minus 1) 2?.

statuscvo [17]

You can use the definition of logarithm and the fact that a positive number raised to any power will always stay bigger than 0.

The domain of the given function is {x | x > 1 and a real number }

The range of the given function is $\mathbb R$ (set of real numbers)

<h3>What is the definition of logarithm?</h3>

If a is raised to power b is resulted as c, then we can rewrite it that b equals to the logarithm of c with base a.

Or, symbolically:

$a^b = c \implies b = log_a(c)$

Since c was the result of a raised to power b, thus, if a was a positive number, then a raised to any power won't go less or equal to zero, thus making c > 0

<h3>How to use this definition to find the domain and range of given function?</h3>

Since log(x-1) is with base 10 (when base of log isn't specified, it is assumed to be with base 10) (when log is written ln, it is log with base e =2.71828.... ) thus, we have a = 10 > 0 thus the input x-1 > 0 too.

Or we have:

x > 1 as the restriction.

Thus domain of the given function is {x | x > 1 and a real number }

Now from domain, we have:

$x > 1\\
x-1 > 0\\
log(x-1) > -\infty\\
log(x-1) + 2 > -\infty\\
f(x) > -\infty$ (log(x-1) > -infinity since log(0) on right side have arbitrary negatively large value which is denoted by -infinity)

Thus, range of given function is whole real number set $\mathbb R$ (since all finite real numbers are bigger than negative infinity)

Thus, the domain of the given function is {x | x > 1 and a real number }

The range of the given function is $\mathbb R$ (set of real numbers

Learn more about domain and range here:

brainly.com/question/12208715

8 0

3 years ago

LeBron goes to the store with six dollars and buys four packs of gum. James goes to the store with five dollars and buys two pac

lilavasa [31]

G = the cost of one pack of gum
LeBron : 6 - 4g
James : 5 - 2g

what they both have after buying gum..
6 - 4g + 5 - 2g = 11 - 6g

7 0

3 years ago

Read 2 more answers

HURRY PLEASE -5(2x − 4)

lana [24]

-10x +20
assuming you want it to equal 0
x would equal 2

3 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago