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Kamila [148]
2 years ago
10

x%5E%7B2%7D" id="TexFormula1" title="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" alt="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Ber [7]2 years ago
5 0
X1= -4, x2= -1/2 is the correct answer
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A(2x+3) = 9x+15+x. Solve for a
zepelin [54]
  • Answer:

<em>a = 5</em>

  • Step-by-step explanation:

<em>a(2x + 3) = 9x + 15 + x</em>

<em>a(2x + 3) = 10x + 15</em>

<em>a(2x + 3) = 5(2x + 3)</em>

<em>a = 5(2x + 3)/(2x + 3)</em>

<em>a = 5</em>

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3 years ago
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Find the sum 7+101/2+14+.........+84​
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Answer: 155.5 Is your sum! (❤´艸`❤)

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3 years ago
Please help this one doesnt nee and explantion
ICE Princess25 [194]

Answer:

$159.99 * .25 "C"

Step-by-step explanation:

5 0
2 years ago
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Graph the image of the given triangle under a dilation with a scale factor of −2 and center of dilation (0, 0) .
Ronch [10]

To dilate an object means to enlarge or reduce the size of the object. The scale factor will determine how much larger or smaller the object will become.  If this factor is greater than 1, the object will increase in size. Otherwise, if the factor is less than 1, the object will decrease in size. So, the dilated object will be similar to its original. On the other hand, when corresponding points of the original and dilated figures are connected by straight lines, the center of dilation is the point where all the lines meet. In this problem, the center is (0, 0). When the center is the origin we need to multiply all the original coordinates of the object by the scale factor given. So:


A(-4, 1) \rightarrow A'=-2(-4,1) \rightarrow A'(8,-2) \\ \\ B(-3, -4) \rightarrow B'=-2(-3,-4) \rightarrow B'(6,8) \\ \\ C(-2, -2) \rightarrow C'=-2(-2,-2) \rightarrow C'(4,4)


So, the graph of the dilated triangle is shown in the Figure below

8 0
3 years ago
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Which point on the number line represents the product of 4 and –2?
Goshia [24]

Answer:

It's A

Step-by-step explanation:

the one above me is incorrect it's "A"

5 0
3 years ago
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