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disa [49]
2 years ago
12

Sheldon needs a pound of meat for a recipe he is planning to make next week. He sets aside a pound of meat and puts it in the fr

eezer for next week. When he takes the meat out of the freezer next week and weighs it, what will he find?(1 point)
He needs more meat for his recipe.
He needs more meat for his recipe.

He has the right amount of meat for his recipe.
He has the right amount of meat for his recipe.

He needs less meat for his recipe.
He needs less meat for his recipe.

He has to replace the spoiled meat for his recipe.
Mathematics
2 answers:
IrinaVladis [17]2 years ago
7 0

Answer:

He needs more

Step-by-step explanation:

padilas [110]2 years ago
6 0

Answer:

He needs to check his work more, and stop asking middle schoolers to do his cooking math.

Step-by-step explanation:

You might be interested in
Please help asap <3
sladkih [1.3K]

Answer:

think it's A (0,6) if not then idk maybe D

6 0
3 years ago
Read 2 more answers
( 7x + 2x - 3x2 + 9x) + (5x + 8x2 - 6x2 + 2x - 5)​
Elena-2011 [213]

Answer:

-x2+25x-5

Step-by-step explanation:

7x+2x−3x2+9x+5x+8x2−6x2+2x−5

=−x2+25x−5

3 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!! Do all real numbers have a decimal expansion? Select each correct answer. Some numbers, such as 1 10 ​
ankoles [38]

Answer:

1. Some numbers, such as \frac{1}{10}, have a decimal expansion that terminates.

2. All real numbers have a decimal expansion.

5. Some numbers, such as \frac{1}{18}, have a decimal expansion that repeats but does not terminate.

Step-by-step explanation:

According to the options, we have,

1. Some numbers, such as \frac{1}{10}, have a decimal expansion that terminates.

It is correct as the decimal expansion of \frac{1}{10} is 0.1, which terminates.

2. All real numbers have a decimal expansion.

This is also correct as both rational and irrational numbers can be written in decimal form.

3. Some numbers, such as \sqrt{13} do not have a decimal expansion

This is not correct as \sqrt{13}=3.605551275 i.e. it has a decimal expansion.

4. Only some real numbers have a decimal expansion.

It is not correct as 2 is correct.

5. Some numbers, such as \frac{1}{18}, have a decimal expansion that repeats but does not terminate.

It is correct as \frac{1}{18}=0.0555555.., which is non-terminating decimal expansion.

So, the correct options are,

1. Some numbers, such as \frac{1}{10}, have a decimal expansion that terminates.

2. All real numbers have a decimal expansion.

5. Some numbers, such as \frac{1}{18}, have a decimal expansion that repeats but does not terminate.

6 0
3 years ago
Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.
Romashka [77]
By definition, the average change of rate is given by:
 AVR =  \frac{f(x2)-f(x1)}{x2-x1}
 We will calculate AVR for each of the functions.
 We have then:

 f(x) = x^2 + 3x interval: [-2, 3]:
 f(-2) = x^2 + 3x  = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18
 AVR = \frac{-2-18}{-2-3}
 AVR = \frac{-20}{-5}
 AVR = 4

 f(x) = 3x - 8 interval: [4, 5]:
 f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7
 AVR = \frac{7-4}{5-4}
 AVR = \frac{3}{1}
 AVR = 3

 f(x) = x^2 - 2x interval: [-3, 4]
 f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8
 AVR = \frac{8-15}{4+3}
 AVR = \frac{-7}{7}
 AVR = -1

 f(x) = x^2 - 5 interval: [-1, 1]
 f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4
 AVR = \frac{-4+4}{1+1}
 AVR = \frac{0}{2}
 AVR = 0


 Answer:
 
these functions from the greatest to the least value based on the average rate of change are:
 f(x) = x^2 + 3x
 
f(x) = 3x - 8
 
f(x) = x^2 - 5
 
f(x) = x^2 - 2x
5 0
3 years ago
Marina walks from the police station at (-10, -4) to her friend Marissa’s house at (3, -8). She stops half way at a fast food re
Volgvan
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -10}}\quad ,&{{ -4}})\quad 
%  (c,d)
&({{ 3}}\quad ,&{{ -8}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{3-10}{2}~~,~~\cfrac{-8-4}{2} \right)
8 0
3 years ago
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