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stepladder [879]
3 years ago
14

How do I write the system of linear equations for this graph?

Mathematics
2 answers:
goblinko [34]3 years ago
5 0

( 1 ) First graph -

x-intercept = 2

y-intercept = -4

y=2x-4

[ Proof ]

2x-4=0\\2x=4\\x=2

For y-intercept, it's y = 0 -4 = -4

( 2 ) Second graph -

x-intercept = 2

y-intercept = 1

slope = -1/2

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

[ Proof ]

-\frac{1}{2}x+1=0\\-x+2=0\\-x=-2\\x=2

For y-intercept, it's y = 0 + 1 = 1

So it's y = 2x - 4

and y = -1/2x + 1

nataly862011 [7]3 years ago
4 0

Answer:y=-2x+0

Step-by-step explanation:

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Can someone help me with this question? You don’t really have to tell me the answer but just how to solve it.
Veronika [31]

part A

if the paperbacks are 40% less, that means 100% - 40% = 60%, they actually are 60% of the price of the hardcover ones.

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{60\% of 24.99}}{\left( \cfrac{60}{100} \right)249.99}\implies 14.994

part B

so, if they return it, they get 1/3 credit, if they try to sell it, is half off, which is more than 1/3.  22 got damaged.

\bf \stackrel{\textit{damaged books at half-price}}{(22\cdot 14.994)\cfrac{1}{2}\implies 164.934}~\hfill \stackrel{\textit{damaged books at one-third price}}{(22\cdot 14.994)\cfrac{1}{3}\implies 109.956} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill 164.934 - 109.956 = 54.978~\hfill

part C

\bf \textit{First Day}~\hfill \stackrel{hardcover}{3(24.99)}+\stackrel{paperback}{8(14.994)}+\stackrel{damaged}{3\left( 14.994\cdot \frac{1}{2} \right)} \implies 217.413 \\\\\\ \textit{Second Day}~\hfill \stackrel{hardcover}{4(24.99)}+\stackrel{paperback}{5(14.994)}+\stackrel{damaged}{8\left( 14.994\cdot \frac{1}{2} \right)}\implies 234.906 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \boxed{17.493}

I don't think she is incorrect, $17.5 is close to $20, she didn't say it was exactly $20, just thereabouts, and 17.5 is around that much.

8 0
3 years ago
4. Find the LCM of the following numbers using prime factors:
Vadim26 [7]

Answer:

The least common multiple of 6, 15 and 40 is <u>120</u>.

The least common multiple of 32, 48 and 72 is <u>288</u><u>.</u>

The LCM of 20 and 24 is <u>120</u>.<em> </em>

The LCM of 30 and 90 is <u>90</u>.

The LCM of 150 is <u>900</u>.

Step-by-step explanation:

<u>Steps to find LCM</u>

  • Find the prime factorization of 150. 150 = 2 × 3 × 5 × 5.
  • Find the prime factorization of 180. 180 = 2 × 2 × 3 × 3 × 5.
  • LCM = 2 × 2 × 3 × 3 × 5 × 5.
  • LCM = 900.
5 0
3 years ago
What is the selling price per widget as a function of the
zalisa [80]

Answer:

1240 widgets

Step-by-step explanation:

<em>The value of the function at 870, the local maximum, is </em>

<em>= -0.02(870)2 + 34.80(870) - 4700 </em>

<em>= -0.02(756900) + 30276 - 4700 </em>

<em>=  -15138 + 25576 </em>

<em>= 10438 </em>

<em>  </em>

<em>So the vertex is (870, 10438) </em>

<em>  </em>

<em>The cost t the company to produce 870 widgets is  </em>

<em>C(870) = 4700 + 5.20(870) = 4700 + 4524 = 9224 </em>

<em>  </em>

<em>So, the cost of the widgets plus the profit must be equal to the total sales, which is divided by the number of widgets reveal their individual price. </em>

<em>(10438 + 9224)/870 = 19662/870 = $22.60 </em>

<em>  </em>

<em>P(x) = 7700 </em>

<em>- 0.02x2 + 34.80x - 4700 = 7700 </em>

<em>-0.02x2 + 34.80x -12400 = 0 </em>

<em>  </em>

<em>x = {-34.80 ± √[(34.80)2 - 4(-0.02)(-12400)]}/2(-0.02) </em>

<em>x = [-34.80 ± √(1211.04 - 992)]/(-0.04) </em>

<em>x = (-34.80 ± √219.04)/(-0.04) </em>

<em>x = (-34.80 ± 14.8)/(-0.04) </em>

<em>x = 870 ± 370 </em>

<em>so, $7700 in profits will be earned at either 500 widgets or </em><em>1240 widgets</em>

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3 years ago
Christopher writes the mixed number as a sum and uses the distributive property.
Zepler [3.9K]

Answer:

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Step-by-step explanation:


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3 years ago
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Kipish [7]

I'm absolutely certain that the answer is 2. (0,-4), (5,3), (-4,3), (3,2) because in a function, x does not repeat

<em>I hope this helps, love!</em>

5 0
3 years ago
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