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3 years ago

How many permutations of the letters in the word REPLICA end in a vowel

Mathematics

1 answer:

Afina-wow [57]3 years ago

8 0

<span>6*5*4*3*2*1*3=2,160
I hope this helps.
</span>

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Please Help.. <br><br>1) Which Choice shows The Values from greatest to least

EleoNora [17]

Answer: C

Step-by-step explanation: 6 being positive will be first, so you can already get rid of A and B. To find the order of the next numbers, know that in negative numbers, the higher the number is, the lower its actual value is.

8 0

2 years ago

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Given the points K(-2;5), L(1,3) and M(-5; 2p). Calculate the value

Advocard [28]

The product of the perpendicular lines is a negative one. Then the value of p will be negative 3.75.

<h3>What is coordinate geometry?</h3>

Coordinate geometry is the study of geometry using the points in space. Using this, it is possible to find the distance between the points, the dividing line is m:n ratio, finding the mid-point of the line, etc.

Given the points K(-2,5), L(1,3) and M(-5, 2p).

If KL is perpendicular to LM. Then we have

The slope of KL will be

KL = (1+2)/(3-5)

KL = -1.5

The slope of LM will be

LM = (1+6)/(3-2p)

LM = 7/(3-2p)

We know the product of the perpendicular lines is a negative one.

KL x ML = -1

-1.5 x 7/(3-2p) = -1

-10.5 = -3 + 2p

2p = -7.5

p = -3.75

More about the coordinate geometry link is given below.

brainly.com/question/1601567

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4 0

1 year ago

Find the Tolal Surface Area

julia-pushkina [17]

The answer is c hope this helped

3 0

2 years ago

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Question 14(Multiple Choice Worth 1 points)

Kryger [21]

Answer: mike traveled 34 miles

3 0

1 year ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

6 months ago