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3 years ago

13

A line passes through the point (6,1) and has a slope of -3.

1 answer:

aleksklad [387]3 years ago

8 0

$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{1})~\hspace{10em} \stackrel{slope}{m}\implies -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{-3}(x-\stackrel{x_1}{6})$

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Determine the holes, vertical asymptotes, and horizontal asymptotes of the rational function y=3x^2+10x-8/x^2+7x+12

kompoz [17]

There would be a hole when x = -4.

If you factor both the top and the bottom, you will see that the factor (x + 4) is in both the numerator and denominator. Therefore, the would be a hole when x = -4 because that would also make the denominator 0.

8 0

3 years ago

In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.

Karo-lina-s [1.5K]

Answer:

$BC=11\ cm$

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

$m\angle DBC=\frac{1}{2}[arc\ DC]$

substitute the values

$60\°=\frac{1}{2}[arc\ DC]$

$120\°=arc\ DC$

$arc\ DC=120\°$

step 2

Find the measure of arc BC

we know that

$arc\ DC+arc\ BC=180\°$ ----> because the diameter BD divide the circle into two equal parts

$120\°+arc\ BC=180\°$

$arc\ BC=180\°-120\°=60\°$

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

$m\angle BDC=\frac{1}{2}[arc\ BC]$

substitute the values

$m\angle BDC=\frac{1}{2}[60\°]$

$m\angle BDC=30\°$

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

$sin(\angle BDC)=BC/BD$

$BC=(BD)sin(\angle BDC)$

substitute the values

$BC=(22)sin(30\°)=11\ cm$

6 0

3 years ago

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).

zzz [600]

Answer:

F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

Step-by-step explanation:

We have given $F(S)=\frac{7e^{-7s}}{s^2-49}$

Now $F(S)=e^{-7s}G(s)$

Here $G(S)=\frac{7}{S^2-49}$

Now first find the Laplace inverse of G(S)

Using partial fraction

$\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}$

$7=A(S-7)+B(S+7)$

On comparing the coefficient

$A=\frac{1}{2}$ and $B=\frac{-1}{2}$

On putting the value of A and B

$G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}$

Taking inverse Laplace

$G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}$

Now in G(s) there is onether term $e^{-7s}$

So F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

6 0

3 years ago

Felix has $100 in the bank and writes a check for $150. He now has

MAXImum [283]

Answer:

Felix has 250 dollars

No change

Lucas has 5 dollars

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Factorize<br><img src="https://tex.z-dn.net/?f=3x%20-%205%20%5Csqrt%7Bx%7D%20%20%3D%202" id="TexFormula1" title="3x - 5 \sqrt{x}

zaharov [31]

Final answer is *drumroll******

x=4

4 0

3 years ago