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SVEN [57.7K]
3 years ago
5

1. Jennifer bought 4 1/2 pounds of hamburger for $22.50. What was the cost per pound?

Mathematics
2 answers:
Mama L [17]3 years ago
5 0

Answer:

.20$

Step-by-step explanation:

4 1/2=4.5

4.5 divided by 22.50=.20

Alex3 years ago
4 0

Answer: Jennifer bought 4 1/2 pounds of hamburger for $22.50.

What was the cost per pound?

.20cents

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Finding the volume of a cube involves multiplying with this formula.

Height * Width * Length = volume

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∛64 = 4

Height = 4

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The only answer choice that makes sense is Option A.

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151

Step-by-step explanation:

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I’m not 100% sure but it could be 225 and 328. Adding those together you get a sum of 553.

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3 years ago
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You own Everything's Coming Up Roses flower shop. Your employee makes 4 deliveries per hour. The distance between the shop and t
qaws [65]
We have 4 deliveries ( 12 1/3 , 8 3/4, 17 2/8 and 23 2/3 miles ) and also the final delivery : 10 5/10 miles = 10 1/2 miles.
We have to add up all the deliveries and to divide the result by 5:
( 12 1/3 + 8 3/4 + 17 1/4 + 23 2/3 + 10 1/2 ) : 5 =
= 72 1/2 : 5 = 72.5 : 5 = 14.5 = 14 1/2
Answer: The average distance for all segments of this trip is 14 1/2 miles. 
4 0
3 years ago
Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
2 years ago
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