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zalisa [80]
2 years ago
8

Can someone plz help TnT

Mathematics
2 answers:
iren2701 [21]2 years ago
7 0
7473739(383)382824(?38!
uranmaximum [27]2 years ago
3 0

Answer:

Slope =-1/3

Y-intercept: (0,4)

Step-by-step explanation:

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Write an equation of a circle that is centered at (1,-4) with a radius of 10.
alexdok [17]

Answer:

(x-1)^2 + (y+4)^2 = 100\\\\

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

Work Shown:

(x-h)^2 + (y-k)^2 = r^2\\\\(x-1)^2 + (y-(-4))^2 = 10^2\\\\(x-1)^2 + (y+4)^2 = 100\\\\

You could expand terms out and simplify, but I think it's more handy to leave it in this form so you can easily spot the center and radius from a glance.

5 0
2 years ago
Read 2 more answers
Graph the function f(x)=3/2x−4. Use the line tool and select two points to graph.
iris [78.8K]

Answer:

Two points on the graph would be (2,-1) and (4,2).

Step-by-step explanation:

You can choose two random x variables such as how i selected 2 and 4. if you change the variable x to those values you can solve for y or in this case f(x).

EX:

3/2*2-4

6/2-4

3-4

f(x)= -1

4 0
3 years ago
Read 2 more answers
Write the equation of the line described in point-slope form. Then, change to Standard Form. Contains points (3/2,-1/2) and (-1/
larisa [96]
Slope point form :
To put in slope point form, label any of the points as either X1,y1 and X and y, then plug in those values into the following equation form.
Y - y1 = m(X-X1)

But before, we must solve for the m value or slope.
M = y2-y1/x2-X1
M = 5/2 - -1/2 / -1/2 - 3/2.
M = 5/2 + 1/2 / -(1/2+3/2)
M = 6/2 / -(4/2)
M = 3/-2

Now we can place it in slope point and also in standard form of a line.

Y-y1 = m(X -X1)

Y - -1/2 = -3/2(X - 3/2)

Y + 1/2 = -3/2(X - 3/2)

This is in slope point form.

Y + 1/2 = -3/2x + 9/4
Y + 1/2 - 1/2 = -3/2x + 9/4 - 1/2
1/2 = 2/4
Y = -3/2x + 7/4
-3/2x = -6/4x
Y = -6/4x + 7/4
Y • 4 = 4( -6/4 X + 7/4)
4y = -6x + 7
4y + 6x = -6x + 6x +7

6x + 4y = 7

This is in standard form. If you have any questions of the steps just ask.
4 0
3 years ago
Help me? idk the answer :P
Alexus [3.1K]
\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\


\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4

\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}

\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9
7 0
3 years ago
What is the slops of the line that passes through the pair of points? (-5.5, 6.1), (-2.5, 3.1)
Svetach [21]

Answer:

most graphs don't go passed 10 so it's undefined

Step-by-step explanation:


8 0
3 years ago
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