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Luda [366]
3 years ago
5

Which equation represents a line that passes through (4,1/3) and has a slope of 3/4?

Mathematics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

\large\boxed{B.\ y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)}

Step-by-step explanation:

The point-slope form of an equation:

y-y_1=m(x-x_1)

m - slope

We have

m=\dfrac{3}{4},\ \left(4,\ \dfrac{1}{3}\right)

Substitute:

y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)

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Some questions have the form y = kx, where k is a number. Drag each question to the correct category below. ( X = y/2 ),( y = .1
fomenos

Answer:

<u>Has the form y = kx</u>

( y = 0.11x)

(y = 0.04x)

<u>Can be put into form y = kx</u>

(z/x = 9)

<u>Other</u>

(5 = xy)

( x = y/2)

Step-by-step explanation:

<u>y = kx</u>

( y = 0.11x)

(y = 0.04x)

<u>kx = y</u>

(z/x = 9)

-----------

( x = y/2 )

( x - 5 = y)

(5 = xy)

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3 years ago
A $240.00 item is marked down by 25%.
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8 0
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How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?
defon
\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1

(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1

\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow  \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2

\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3

\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}



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3 years ago
Equation for elasticity
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5 0
3 years ago
Screenshot of the question attatched. please help me
Law Incorporation [45]
<h2>Rules of Exponents</h2>

There are many rules that apply to operations involving exponents. In this question, we'll need the following:

<h3>The division rule</h3>
  • ⇒ \dfrac{a^n}{a^m}=a^{n-m}

<h2>Solving the Question</h2>

We're given:

z^{21}\div z^7=z^k (solve for k.)

⇒ Rewrite as a fraction:

\dfrac{z^{21}}{z^7}=z^k

⇒ Use the division rule:

z^{21-7}=z^k\\z^{14}=z^k\\14=k

Therefore, the value of <em>k</em> is 14.

<h2>Answer</h2>

<em>k</em> = 14

5 0
2 years ago
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