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choli [55]
3 years ago
11

Graph the system of equations. {4x+4y=16 x+6y=−6 Use the Line tool to graph the lines.

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

4x+4y=16 -----> equation A

x+6y=-6 -----> equation B

we know that

The solution of the system of equations is the intersection point both graphs

In this problem

The intersection point is (6,-2)

therefore

The solution is the point (6,-2)

using a graphing tool

see the attached figure

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Plz with steps .. it's very hard can anyone plz
liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

7 0
2 years ago
Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.
svetoff [14.1K]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
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 % template detailing
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\

\bf \bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

now, with that template in mind, let's see

\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}

so, just change C to +3, thus C/B is 3/1
6 0
3 years ago
Please answer the 2 question <br> l will mark as brainiest
N76 [4]

Answer:

1. A=πr2=π·4.62≈66.4761

A≈66.48

2. A=πr2=π·22≈12.56637

A≈12.57

please mark as brainiest

3 0
3 years ago
Read 2 more answers
​ F(1) =-3 f(1)=−3 f(n)=2⋅f(n−1)+1 ​ f(2)=
bogdanovich [222]

Answer:

3

Step-by-step explanation:

f(1)=−3

f(n)=2⋅f(n−1)+1

​ f(2)= 2×1(2-1)+1

       = 2×1+1

       = 2+1

        = 3

7 0
3 years ago
Read 2 more answers
Write an equation of the line passing through point p that is perpendicular to the given line. p(3,1), y=13x−5
Anarel [89]
You know two lines y=ax+b and y=mx+n that are perpendicular, so we have the product a*m=-1

+ This line is perpendicular to y=13x-5, so it has equation: y=-1/13x+b
+ And it passes through the point (3;1), so we have x=3, y=1. So 1=-1/13 *3+b
and b= 1+3/13= 16/13
And we have y=-1/13x+16/13
Have fun
7 0
3 years ago
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