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3 years ago

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

3 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:

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