Find the distance between the points (-4, -5 and (3, -1.

2 answers:

7 0

≡ We know that:
⇔ $(x_{1},y_{1})$ → $(-4,-5)$
⇔ $(x_{2},y_{2})$ → $(3,-1)$

≡ Solution:
⇒ $\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}$
⇒ $\sqrt{[(3-(-4))^{2}+(-1-(-5))^{2}]}$
⇒ $\sqrt{[7^{2}+4^{2}]}$
⇒ $\sqrt{[49+16]}$
⇒ $\boxed{\sqrt{65}}$

tatyana61 [14]2 years ago

7 0

Answer: $\sqrt{65} \\$

i got it right in my quiz

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3y-15-6x in slope intercept form

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2 years ago

Can someone help please

myrzilka [38]

<h3>Answer: 15x^(7/3) - 8x^(7/4) + x + 9000</h3>

=========================================================

Explanation:

If you know the cost function C(x), to find the marginal cost, we apply the derivative.

Marginal cost = derivative of cost function

Marginal cost = C ' (x)

Since we're given the marginal cost, we'll apply the antiderivative (aka integral) to figure out what C(x) is. This reverses the process described above.

$\text{Cost} = \text{antiderivative of marginal cost}\\\\\displaystyle C(x) = \int \left(35x^{4/3} - 14x^{3/4} + 1\right)dx\\\\$

$C(x) = \frac{1}{1+4/3}*35x^{4/3+1} - \frac{1}{1+3/4}*14x^{3/4+1} + x + D\\\\C(x) = \frac{1}{7/3}*35x^{7/3} - \frac{1}{7/4}*14x^{7/4} + x + D\\\\C(x) = \frac{3}{7}*35x^{7/3} - \frac{4}{7}*14x^{7/4} + x + D\\\\C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\$

D represents a fixed constant. I would have used C as the constant of integration, but it's already taken by the cost function C(x).

To determine the value of D, we plug in x = 0 and C(x) = 9000. This is because we're told the fixed costs are $9000. This means that when x = 0 units are made, you still have $9000 in costs to pay. This is the initial value. You'll find that all of this leads to D = 9000 because everything else zeros out.

Therefore, we go from this

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