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Paladinen [302]
3 years ago
6

3x to the power of 2 - 363 = 0

Mathematics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

yes

Step-by-step explanation:

because it would be 300 so u would be in negative so ya

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21. The parent function of the following graph is f(x) = 2^x. What is the equation of the following graph?
Scorpion4ik [409]

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5

Step-by-step explanation:

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A fashion designer makes and sells hats. The material for each hat cost $5.50. The Hats sell for $12.50 each. The designer spend
labwork [276]

The designer must sell 200 hats for breaking even.

Step-by-step explanation:

Given,

Cost for material for each hat = $5.50

Selling price of each hat = $12.50

Cost spend on advertising = $1400

Let,

x represents the number of hats.

Revenue = 12.50x

Costs = 5.50x+1400

For breaking even,

Revenue = Cost

12.50x=5.50x+1400\\12.50x-5.50x=1400\\7x=1400

Dividing both sides by 7

\frac{7x}{7}=\frac{1400}{7}\\x=200

The designer must sell 200 hats for breaking even.

Keywords: addition, division

Learn more about division at:

  • brainly.com/question/916181
  • brainly.com/question/9136949

#LearnwithBrainly

7 0
3 years ago
This box is a right rectangular prism with a base that has dimensions of 6 m and 7 m. The volume of the box is 378 m3. What is t
Over [174]

Answer:

29.07m³

Step-by-step explanation:

7+6=13

378÷13=29.07m³

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What is the solution of the system? Use either the substitution method or the elimination method. 4x + 2y = 104x + 2y = 10 x − y
bixtya [17]
4x+2y=10 Equation 1
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Solving by substitution method.
Isolate x from equation 2.
x=y+13
Substitute value of x in equation 1
4(y+13)+2y=10
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y=-7
Now substitute value of y in x=y+13
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8 0
3 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
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