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zhannawk [14.2K]
3 years ago
9

Which expression is equivalent to the following 1/a^5?

Mathematics
1 answer:
Fofino [41]3 years ago
7 0

Answer:

if this is a fraction, then negative 5 over a⁶

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This is the problem on my grandson's homework
Charra [1.4K]
Choose letters to  represent the unknowns:  x and y.  We are told that x-y = 361.

We could choose any number for x and then calculate y:

Suppose we let x = 500.  Then 500 - y = 361, and y = 139.

Suppose we let x = 1.  Then 1 - y = 361, and y = -360.

And so on.

3 0
4 years ago
Ive been stuck with this for a while now please help
Makovka662 [10]

To write 2^7, you would do 2*2*2*2*2*2*2 because your multiplying 2 by itself 7 times. The only factor is, 2. Now try and do the rest, if you still need help, comment and I'll edit my answer and give you more hints. you got this. I believe in you :)

5 0
3 years ago
I cannot find the answer to this can someone please help me
dlinn [17]

Answer:

A = 1,144 ft²

Step-by-step explanation:

The base of the quadrilateral is also x.  We know this since there are 2 90° angles, there fore the other 2 angles are also 90°, making the top and bottom sides of the quadrilateral parallel to each other, and equal in measure.  

Since the entire base of the yard is 2x, the base of the triangle is x.  

The area of the triangle is

A = (1/2)bh      where b is the base, and h is the height

For this example...

A = (1/2)x(44)                  (x is the base, and 44 is the height)

So

A = (1/2)(44)x

   A = 22x

The area of the quadrilateral is

A = LW    

A = 44x

The total area of the yard is

A = 22x + 44x

  A = 66x

We are told that the entire area is 3,462. so

3,462 = 66x           now solve for 'x'

3,462/66 = x           (divide both sides by 66)

52 = x        

So the area of the picnic tables is the area of the triangle, which is

A = 22(52) = 1,144 ft²

5 0
4 years ago
Exercise 3.7.4: let a = 2 1 0 0 2 0 0 0 2 .
swat32

With

\mathbf A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}

we have

\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&0&2-\lambda\end{vmatrix}=(2-\lambda)^3

so \mathbf A has one eigenvalue, \lambda=2, with multiplicity 3.

In order for \mathbf A to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But \mathbf A-2\mathbf I has nullspace of dimension 2, since

\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\mathbf 0\implies x=0\text{ or }y=0

That is, we can only obtain 2 eigenvectors,

\begin{bmatrix}1\\0\\0\end{bmatrix}\text{ and }\begin{bmatrix}0\\0\\1\end{bmatrix}

and there is no other. We needed 3 in order to complete the basis of eigenvectors.

3 0
3 years ago
Please help confused
masya89 [10]

Answer:

option C is the answer.

6 0
3 years ago
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