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3 years ago

Which of these is also (always) a rhombus? (Select all that apply.)

Mathematics

2 answers:

Snezhnost [94]3 years ago

3 0

Answer:

A square and maybe a kite.

Step-by-step explanation:

Assoli18 [71]3 years ago

3 0

Answer:

D. Square

Step-by-step explanation:

Here is a good images that explains what is what:

Read it from bottom to top. It cannot be reversed.

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In which quadrant is the point (-3,5) located?

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Answer:

(-3,5) lies in Quadrant 2

the solution is b

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3 years ago

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What is the number of fluid ounces in 7 cups.

vovikov84 [41]

Answer:

56 ounces :)

Step-by-step explanation:

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3 years ago

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Will mark BRAINLIEST!!!!

Virty [35]

Answer:

The first one is your answer.

Step-by-step explanation:

What you do is you have to look at the 2/18 in the square root. That can come out to be 1/3. So you know that a 3 is going to be on the bottom.

Now you look at the x^5. What you do is you make it until you have 5 x's like this x, x, x, x, x. You then put them in pairs of two.

x,x x,x x. There are two pairs of two so that comes out.

So you have 1x^2sqrtx/3. which is the first one.

#1 is your answer.

6 0

3 years ago

Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of

VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are $26^{5} = 11,881,336$ of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

$26*25*24*23*22 = 7,893,600$

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

$T = T_{1} + T_{2} + T_{3}$

$T_{1}$ is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

$T_{1} = 26^{3}*25 = 439,400$

$T_{2}$ is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

$T_{2} = 26^{3}*25 = 439,400$

$T_{3}$ is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

$T_{3} = 26^{3} = 17,576$

$T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376$

896,376 of these words start with an a or end with a z or both

4 0

3 years ago

F=e−yi−xe−yj is conservative. find a scalar potential f and evaluate the line integral over any smooth path c connecting a(0,0)

Alexxx [7]

If $\mathbf F$ is conservative, then there is a scalar function $f$ such that

$\nabla f(x,y)=\mathbf F(x,y)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=e^{-y}\,\mathbf i-xe^{-y}\,\mathbf j$

Setting the first components equal to one another, we can integrate both sides to find

$\dfrac{\partial f}{\partial x}=e^{-y}\implies f(x,y)=xe^{-y}+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=-xe^{-y}+\dfrac{\mathrm dg}{\mathrm dy}=-xe^{-y}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0$
$\implies g(y)=C$

so that

$f(x,y)=xe^{-y}+C$

By the fundamental theorem of calculus, we have that

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(1,1)-f(0,0)=\frac1e$

4 0

3 years ago