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Kamila [148]
3 years ago
10

A 22-ounce bowl of crackers contains 18 percent cheese crackers. Aida computed 18 percent of 22. Her work is shown below.

Mathematics
2 answers:
Degger [83]3 years ago
3 0
Aida did not use 0.18 for 18%.

When you are finding the percent of something, you can also do this in the attachment below. Yes it is written on a napkin. Hope I helped.

Tamiku [17]3 years ago
3 0

Answer:

Your answer to this question is A.

Step-by-step explanation:

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Lady_Fox [76]

Answer: answer b

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Rewrite in simplest terms: (-8x + 5) - (x – 7)​
Ilia_Sergeevich [38]

Answer:

-9x+12

Step-by-step explanation:

(-8x + 5) - (x – 7)​

Distribute the minus sign

(-8x + 5) - x + 7

Combine like terms

(-8x -x+ 5+7) ​

-9x +12

7 0
3 years ago
Read 2 more answers
WORTH 30 PTS
kifflom [539]

Answer:

Done

Step-by-step explanation:

8 0
3 years ago
Sin5A/SinA-cos5A/cosA=4cos2A​
irina1246 [14]

Answer:

See Explanation

Step-by-step explanation:

\frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ LHS = \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}   \\  \\  =  \frac{ \sin5A \:\cos A -  \cos5A \:  \sin A}{\sin A \:\cos A }  \\  \\  =  \frac{ \sin(5A -A )}{\sin A \:\cos A}  \\  \\ =  \frac{ \sin 4A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =4\cos 2A \\  \\  = RHS \\  \\ thus \\  \\  \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ hence \: proved

5 0
3 years ago
If 1900 square centimeters of material are available to make a box with a square base and an open top, find the largest possible
Evgen [1.6K]

Answer:

Volume = 7969 cubic centimeter

Step-by-step explanation:

Let the length of each side of the base of the box  be A and the height of the box be H.

Area of material required to make the box  is equal to  is A^2 + 4*A*H.

A^2 + 4*A*H = 1900

 Rearranging the above equation, we get -  

`H = \frac{(1900 - A^2)}{(4*A)}

Volume of box is equal to product of base area of box and the height of the box -  

V = A*A* H

Substituting the given area we get -

\frac{A^2*(1900 - A^2)}{4A} = \frac{(1900*A - A^3)}{4}

For maximum volume

\frac{dV}{dA} =0

\frac{ 1900}{4} - \frac{3*A^2}{4} = 0

A^2 = \frac{1900}{3}

Volume of the box

= \frac{\frac{1900}{3}*(1900 - \frac{1900}{3}) }{4 * \sqrt{\frac{1900}{3} } }

= 7969 cubic centimeter

3 0
3 years ago
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