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choli [55]
4 years ago
12

Show all work to write the equations of the lines, representing the following conditions, in the form y = mx + b, where m is the

slope and b is the y-intercept: Part A: Passes through (−2, 2) and parallel to 4x − 3y − 7 = 0 (2 points) Part B: Passes through (−2, 2) and perpendicular to 4x − 3y − 7 = 0 (2 points)
Mathematics
1 answer:
Basile [38]4 years ago
5 0
4x - 3y - 7 = 0.....put this in y = mx + b form
-3y = -4x + 7
y = 4/3x - 7/3.....slope here is 4/3. 
=======================
A parallel line will have the same slope.

y = mx + b
slope(m) = 4/3
(-2,2)....x = -2 and y = 2
now sub into the formula and find b, the y int
2 = 4/3(-2) + b
2 = - 8/3 + b
2 + 8/3 = b
6/3 + 8/3 = b
14/3 = b
so ur parallel equation is : y = 4/3x + 14/3 <==
========================
slope = 4/3. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, flip the original slope and change the sign. So we have 4/3.....now we flip it making it 3/4....now we change the sign, making it -3/4. So our perpendicular line will have a slope of -3/4.

y = mx + b
slope(m) = -3/4
(-2,2)...x = -2 and y = 2
now we sub into the formula and find b, the y int
2 = -3/4(-2) + b
2 = 3/2 + b
2 - 3/2 = b
4/2 - 3/2 = b
1/2 = b
so ur perpendicular equation is : y = -3/4x + 1/2 <===


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It will take  <u><em>80 days</em></u>  for the bull calf to reach a weight of 500 kilograms.

Step-by-step explanation:

Given:

The weight of a bull calf is 388 kilograms.

Now, to find the weight of bull calf of how long it will take to reach a weight of 500 kilograms, if it’s weight increases at a rate of 1 2/5 kilograms per day.

Required weight which to be increased = 500 - 388 = 112 kilograms.

Rate of weight increase = 1\frac{2}{5}=\frac{7}{5}

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Thus, the time required = \frac{required\ weight}{rate\ of\ weight}

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<em>The time required   =    80 days</em>.

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\approx 9.0\:\mathrm{units}

Step-by-step explanation:

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Step-by-step explanation:

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myrzilka [38]
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=========================================================

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Marginal cost = derivative of cost function

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C(x) = 15x^{7/3} - 8x^{7/4} + x + 9000\\\\

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