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2 years ago

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Determine whether the system of equations has one solution, no solution, or infinitely many solutions by graphing. If the system

has one solution, name it.
3x-y = 14
-6+ 2y = -28

A. One solution (2,2)
B. Infinitely many solutions
C. No solutions
D. No Solutions

2 answers:

Svetlanka [38]2 years ago

8 0

One solution

I’m pretty sure that’s right

Juli2301 [7.4K]2 years ago

5 0

Answer:

One solution

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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What’s the equation for this

LekaFEV [45]

<h2>Answer:</h2>

Hope it work for u

<h2>Step-by-step explanation:</h2>

We can make points from the given figure as ,

1 point =(x1, y1)=(-2, -3)

2nd point =(x2, y2)= (0, -6)

3rd point = (x3, y3)=(2 , -9)

Now the equation of line is,

<h2>Y=m X+ c where m is slope and c is constant </h2>

now two point formula of slope is ,

<h2>m = y2-y1/x2-x1</h2>

putting points in it we get

m=(-6-(-3))/(0-(-2)

m=-3/2 is slope

Now to find C putt the value of m and 3rd point in equation of line we get

y3=m(x3)+c

-9=(-3/2)(2) +c

C= -9+3

C=-6 is value of c

Now at last putt the value of C and m in equation of line we get the required equation ,

<h2><u>Y=(-3/2)X-6 </u> Ans</h2>

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2 years ago

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Sadie has $1 worth of nickels and dimes. She has twice as many nickels as dimes. Write a system of equations that could be used

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N + 2d =1
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Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

$\bf \begin{array}{llll}
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

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2 years ago