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2 years ago

Using the midpoint formula find BC if B(8,-7) and C(-4,-2)

Mathematics

1 answer:

arsen [322]2 years ago

6 0

Answer:

(2, -4.5)

Step-by-step explanation:

Used midpoint formula

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A store pays $35 for a fish tank. The markup is 20%. What is the selling price

snow_lady [41]

The selling price is $42

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3 years ago

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The mean points obtained in an aptitude examination is 111 points with a variance of 400.

suter [353]

Answer:

0.9128 = 91.28% probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled

Step-by-step explanation:

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3 years ago

3X+7=X+25 SOLVE FOR X. A.9 B.25 C.34 68.

Ipatiy [6.2K]

3 x 9 + 7 = 34
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3 years ago

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Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont

Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

$f(x) = x^{3}-9\cdot x^{2}+ 3$ , $2 \leq x \leq 12$

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

$f'(x) = 3\cdot x^{2}-18\cdot x$

$3\cdot x^{2}-18\cdot x = 0$

$3\cdot x \cdot (x-6) = 0$ (Eq. 1)

As we can see, only a solution is a valid critical value. That is: $x = 6$

Second, we determine the second derivative formula and evaluate it at the only critical point:

$f''(x) = 6\cdot x -18$ (Eq. 2)

x = 6

$f''(6) = 6\cdot (6)-18$

$f''(6) =18$ (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

$f(2) = 2^{3}-9\cdot (2)^{2}+3$

$f(2) = -25$

x = 6

$f(6) = 6^{3}-9\cdot (6)^{2}+3$

$f(6) = -105$

x = 12

$f(12) = 12^{3}-9\cdot (12)^{2}+3$

$f(12) = 435$

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

6 0

3 years ago

Inverse laplace of L^-1 {5s/s^2 + 3s - 4}

Salsk061 [2.6K]

Answer:

$e^t+e^{-4t}$

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

$\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\$

$From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}$

5 0

3 years ago