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jok3333 [9.3K]
3 years ago
11

Need help with 9(c) plzzz

Mathematics
1 answer:
MAVERICK [17]3 years ago
5 0
Since we are given the values for both rows only in the second column, we can use this to solve for the rest of the missing values. Simply divide 3 by 2.49 and multiply that resultant by and multiply that by any missing value for cookies to find the missing cost. In order to solve for the cookies when cost is given, divide 2.49 by 3 and multiply that resultant. I will solve for the first, third, and fourth columns.

3/29=1.2
For column 1, 1.2*1=1.20
For column 3, 1.2*20=24.10
For column 4, 1.2*100=12.00

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GET BRAINLIEST FOR THE CORRECT ANSWER AND EXPLANATION!
Tresset [83]

Interpreting the inequality, it is found that the correct option is given by F.

------------------

  • The first equation is of the line.
  • The equal sign is present in the inequality, which means that the line is not dashed, which removes option G.

In standard form, the equation of the line is:

x + 2y = 6

2y = 6 - x

y = -0.5x + 2

Thus it is a decreasing line, which removes options J.

  • We are interested in the region on the plane below the line, that is, less than the line, which removes option H.

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  • As for the second equation, the normalized equation is:

3x^2 + 3y^2 = 12

3(x^2 + y^2) = 12

x^2 + y^2 = 4

  • Thus, a circle centered at the origin and with radius 2.
  • Now, we have to check if the line x + 2y - 6 = 0, with coefficients a = 1, b = 2, c = -6, intersects the circle, of centre x = 0, y = 0
  • First, we find the following distance:

d = \sqrt{\frac{|ax + by + c|}{a^2 + b^2}}

  • Considering the coefficients of the line and the center of the circle.

d = \sqrt{\frac{|1(0) + 2(0) - 6|}{1^2 + 2^2}} = \sqrt{\frac{6}{5}} = 1.1

  • This distance is less than the radius, thus, the line intersects the circle, which removes option K, and states that the correct option is given by F.

A similar problem is given at brainly.com/question/16505684

3 0
3 years ago
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Answer:

See solutions for detail.

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The volume's rate of change is written as a function of time.

-\frac{dh}{dt} is the rate of change in the height of water in the tank with respect to time, t.

b.  \frac{dV}{dt}- is the only constant. Water flows into the constant at a constant rate, say 6cm^3 per minute.

c. \frac{dV}{dt} is positive. Volume water in the take  is increasing from time to time.

-The volume at time t=1 is greater than the volume at t=0, hence, it's a positive rate of change.

d. \frac{dh}{dt} is a positive rate. The initial height of water in the tank is zero.

-The final height at time t is 0.25h. The height is increasing with time.

Hence, it is positive.

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