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3 years ago

It costs a ski resort $15/hour to hire one front-loader operator. If the ski resort plans to

Mathematics

1 answer:

Mamont248 [21]3 years ago

4 0

Answer:

$720

Explanation:

3 (yd³) per minute per worker.

4*3=12 yd³ per minute in total.

12*60 = 720 yd³ per hour.

8920 yd³ ÷ 720 yd³ = 12.388889 hours of work.

15$ a worker/hour x 4 workers = $60 per hour in total. Since there are only 12 whole hours in 12.388889, $60/hour x 12 full hours = $720 for all work done.

You might be interested in

A fair dice is rolled. Work out the probability of getting a factor of 12.

Pavlova-9 [17]

Answer:

P (factors of 12) = 5/6

Step-by-step explanation:

The possibilities are 1,2,3,4,5,6

Factors of 12 are 1,2,3,4,6,12

So out of the possibilities rolled

1,2,3,4,6 are factors of 12 (5 of the 6)

P (factors of 12) = 5/6

8 0

4 years ago

Read 2 more answers

Type the correct answer in each box. Use numerals instead of words.

VARVARA [1.3K]

Answer:

see explanation

Step-by-step explanation:

f(x) = - (x + 1)(x - 3)(x + 2)

to find the zeros let f(x) = 0 , that is

- (x + 1)(x - 3)(x + 2) = 0

equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

x - 3 = 0 ⇒ x = 3

x + 2 = 0 ⇒ x = - 2

the zeros are x = -1, 3 and x = - 2

to find the y- intercept let x = 0

f(0) = - (0 + 1)(0 - 3)(0 + 2) = - (1)(- 3)(2) = - (- 6) = 6

y- intercept is located at (0, 6 )

6 0

2 years ago

The table shows the number of 7th- and 8th-grade students who belong to an after-school club

netineya [11]

Answer: whats the question though

Step-by-step explanation:

7 0

3 years ago

Select all statements below which are true for all invertible n×n matrices A and B

Mekhanik [1.2K]

Answer:

a. False

b. False

c. True

d. True

e. False

f. False

Step-by-step explanation:

Hi,

We have certain properties for matrices,<em> (where A and B are nxn matrices and I is the identity matrix) </em>:

${(A^{-1})}^{-1} = A$

$(AB)^{-1} = B^{-1}A^{-1}$

$(A')^{-1} = (A^{-1})'$

$(A^{n})^{-1} = (A^{-1})^{n} = A^{-n}$

$AA^{-1} = A^{-1}A = I$

$AI = IA = A$

Using these properties, we verify the provided statements:

A. False.

None of the properties help verify this statement. We ca use an example for counter:

Let $A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right]$ and $B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right]$ , we calculate the L.H.S:

$A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right]$

The square of (A+B):

$(A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right]$

Lets calculate the R.H.S:

$A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3) &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right]$

$A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right]$

This proves that: L.H.S ≠ R.H.S

Hence, A is false.

B. False

This can only hold when the eigenvalues for A are real.

$trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}$

C. True

This is a simplification of the distribution property of matrices.

D. True

The property that inverse is possible for any "n" value of the matrix.

E. False

Similar to part A, we can show that this property is invalid for any nxn matrix. Let:

$A = \left[\begin{array}{cc}1&2\\0&1\end{array}\right] \\A^{-1} = \left[\begin{array}{cc}1&-2\\0&1\end{array}\right]$

L.H.S:

$A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right] = \left[\begin{array}{cc}2&0\\0&2\end{array}\right]$

$(A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right]$

R.H.S:

$A^{9} = \left[\begin{array}{cc}1&512\\0&1\end{array}\right] \\A^{-9}= \left[\begin{array}{cc}1&-0.001953125\\0&1\end{array}\right]\\\\\\A^{9} + A^{-9} = \left[\begin{array}{cc}2&512\\0&2\end{array}\right] \\$

Since, L.H.S ≠ R.H.S, the statement is false.

F. False

This is a basic matrix rule, that commutative property does not apply on matrices.

3 0

3 years ago

Phillip bought apples from the grocery for $2.20 per pound. If he bought 2.5 pounds of apples on Monday and 1.2 pounds of apples

vagabundo [1.1K]

Monday, 2.5×2.2=5.50
Tues, 1.2×2.2=2.64

5.50+2.64=8.14

He spent $8.14 total

6 0

4 years ago