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Zinaida [17]
3 years ago
10

Given the formula:

" title="x = \frac{1 + a}{1 - a} " alt="x = \frac{1 + a}{1 - a} " align="absmiddle" class="latex-formula">
Solve for "a" when x=12.​
Mathematics
1 answer:
shtirl [24]3 years ago
8 0

12 = \frac{1+a}{1-a} => 12(1-a)= 1+ a \\=> 12 -12a = 1 + a => 12a+1 = 12-1 \\=> 13a = 11 => a= \frac{11}{13}

ok done. Thank to me :>

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Graph the equation. Let x​3, ​2, ​1, ​0, 1,​ 2, and 3. y=3x-1
spin [16.1K]

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

The attachment shows the table of point values and the graph.

4 0
3 years ago
Help me please with this question
Natalka [10]

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Answer:

  c.  3 to 4

Step-by-step explanation:

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8 0
3 years ago
Write the equation of the circle with center (3, 2) and (−6, −4) a point on the circle.
yulyashka [42]
The equation for a circle is as followed:

(x-h)^2+(y-k)^2=r^2

where the center of the circle is at (h,k) and the radius of the circle is r.

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Plug in the two points:

d= \sqrt{(3-(-6))^2+(2-(-4))^2}

d= \sqrt{(9)^2+(6)^2}

d= \sqrt{117}

If the distance from the center to the edge of the circle is the square root of 117, then r^2 = 117.

The answer is:

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6 0
3 years ago
Ana has $500 to spend on shopping for her trip. She spends her money on the following: $120 on swimsuits$50 on towels$100 on clo
shtirl [24]
295 she spent
y dollars she can spend on other expanses
then 295+x+y=500
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4 0
3 years ago
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Thepotemich [5.8K]
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ok so

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n-1=k

rCk times (a)^{r-k}(-b)^k
and rCk=\frac{r!}{k!(r-k)!}

so

4th term
4-1=3
6 is exponent

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(20)(27)(-8)(x^3)(y^3)=-4320x^3y^3



the 4th term is -4320x^3y^3

7 0
3 years ago
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