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ratelena [41]
3 years ago
13

The sum of 9 and twice a number

Mathematics
1 answer:
Marrrta [24]3 years ago
4 0
9 + n = x ?
i'm guessing since there's no answer to what the sum is
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Fill in the blanks thanks
Mademuasel [1]

Answer:

Um threre is no blanks or there is no question.

Step-by-step explanation:

5 0
3 years ago
Simplify the expression: 2(4 + 6f) plz help this is my end of the year assesment Thanks!
julia-pushkina [17]

Answer:

8+12f

hope this helps

have a good day :)

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs
LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

h=\frac{400}{\pi r^2}

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: h=\frac{400}{\pi r^2}

Therefore:

C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})

C(r)=0.12\pi r^2+\frac{24}{r}

C(r)=\frac{0.12\pi r^3+24}{r}

The minimum cost occurs when the derivative of the Cost =0.

C^{'}(r)=\frac{6\pi r^3-600}{25r^2}

6\pi r^3-600=0

6\pi r^3=600

\pi r^3=100

r^3=\frac{100}{\pi}

r^3=31.83

r=3.17 cm

Recall that:

h=\frac{400}{\pi r^2}

h=\frac{400}{\pi *3.17^2}

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0
3 years ago
Explain why the two figures below are not similar. Use complete sentences and provide evidence to support
Andreas93 [3]

Answer:

they are not like terms

Step-by-step explanation:

8 0
3 years ago
What is a the positive solution of x^2 - 36 =5x​
Minchanka [31]

Answer:

<h2>x = 9</h2>

Step-by-step explanation:

x^2-36=5x\qquad\text{subtract}\ 5x\ \text{from bot sides}\\\\x^2-5x-36=0\\\\x^2+4x-9x-36=0\qquad\text{distribute}\\\\x(x+4)-9(x+4)=0\\\\(x+4)(x-9)=0\iff x+4=0\ \vee\ x-9=0\\\\x+4=0\qquad\text{subtract 4 from both sides}\\\\x=-40

3 0
3 years ago
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