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Morgarella [4.7K]
2 years ago
5

Someone Help me PLS ...

Mathematics
1 answer:
zavuch27 [327]2 years ago
8 0

Answer: Number 1 would be F

Number 2 would be C

Number 3 is D

Number 4 is B

Number 5 is A

Number 6 is E

Step-by-step explanation:

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I need help with this ASAP!
earnstyle [38]

Answer and Step-by-step explanation:

1) If you graph the lines, you will get the lines in the first attachment. You will see that the intersection point is (3, 5), which is the solution to that system.

2) First, let's convert both of these into slope-intercept form so that they're easier to graph.

x - 3y = 2

x = 3y + 2

3y = x - 2

y = \frac{1}{3}x-\frac{2}{3}

AND

-3x + 9y = -6

9y = 3x - 6

y = \frac{3}{9} x-\frac{6}{9} =\frac{1}{3} x-\frac{2}{3}

We see that these two lines are exactly the same, which means that no matter what coordinate works for one, it will work for the other. In other words, there are infinitely many solutions.

And, if you graph these lines, you will get the graph in the second attachment, where they are the same line.

Hope this helps!

5 0
3 years ago
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The M & P Pencil company that makes mechanical pencils expects to have two defective pencils for every 15 made. What is the
sattari [20]
598 defective pencils
7 0
3 years ago
Simplify -2 1/9 - (-4 1/9)
ZanzabumX [31]

Answer:

2

Step-by-step explanation:

The easiest thing is to convert it to an improper fraction.

-2 1/9 turns into -19/9. -4 1/9 turns into -37/9.

-19/9 - (-37/9) here you would add because you are subtracting a negative number.

-19/9 + 37/9 change it to positive and then add.

-19+37=18

it was 18/9. that simplified is 2.

4 0
2 years ago
Please help me... with at least 3 or 4.
EleoNora [17]

3.right angle
4. Right angle
5 0
2 years ago
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I need help with #11
Troyanec [42]
\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}
\\\\
\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}\\\\
\end{array}

\bf \begin{array}{llll}


\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, with that template in mind, let's take a peek at this function

\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically
8 0
3 years ago
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