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Setler [38]
3 years ago
11

What is 9.36•10~4 in standard form

Mathematics
1 answer:
marta [7]3 years ago
8 0
With exponents of 10, the exponent really just tells you how many zeroes there are behind the 1. So, for example, 10^4=10000 (4 zeroes behind 1). When multiplying decimals by exponents of 10, you move the decimal to the right, and the amount of places is the exponent number. For example, if I multiply 5.34 by 10^2, I move the decimal over to the right 2 times and get 534.

So, if you just multiply these you get 9.36*10000, which is 93600.
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It cost $120 to buy the materials for crown molding in the completed office. (Crown molding is the decorative border where the w
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4 0
2 years ago
Which solid has a greater volume? O A. Figure A 48 B. Figure B O C. they are equal A 2 O D. need more information​
MatroZZZ [7]

9514 1404 393

Answer:

  B.  Figure B

Step-by-step explanation:

The figure is difficult to read. We assume the height of the pyramid is 9, and the radius of the cylinder is 1.

The pyramid volume is ...

  V = 1/3Bh

  V = 1/3(4²)(9) = 48 . . . cubic units

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The cylinder volume is ...

  V = πr²h

  V = π(1²)(48) = 48π . . . cubic units

The cylinder has π times as much volume as the pyramid. Figure B is larger.

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<em>Additional comment</em>

If the diameter (not the radius) of the cylinder is 1 unit, then its volume is 12π cubic units and the pyramid has more volume.

7 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
Help what’s the answer
Anni [7]
The answer to this one is the 3rd one
3 0
2 years ago
Read 2 more answers
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