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alexgriva [62]
2 years ago
7

Help need by tomorrow plzzz

Mathematics
1 answer:
aliya0001 [1]2 years ago
3 0
11.) 1.5
12.)17 1/2
13.)543
533+10
( Just remember absolute value.)
14.)$ 75.15

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Match the circle equations in general form with their corresponding equations in standard form. Not all will be used. 
Xelga [282]
<span>The standard form of the equation of a circumference is given by the following expression:

</span>(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius
<span>
On the other hand, the general form is given as follows:

</span>x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}<span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span>\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46

Equations in General Form:

\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0

\bold{6)} \ x^{2}+y^{2}+2x-12y

So let's match each equation:

\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4

Then, its general form is:

x^{2}+y^{2}-12x-8y-4=0

<em><u>First. a) matches 5) </u></em>

\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20

Then, its general form is:

x^{2}+y^{2}-4x+12y-20=0

<em><u>Second. b) matches 1) </u></em>

\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5

Then, its general form is:

x^{2}+y^{2}+4x+6y-5=0

<em><u>Third. c) matches 3)</u></em>

\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9

Then, its general form is: x^{2}+y^{2}+2x-12y-9=0

<em><u>Fourth. d) matches 6)</u></em>
6 0
3 years ago
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Using transformations
Delvig [45]
I have the graphed answer right here

4 0
2 years ago
What is the sum of the finite arithmetic series? 26 + 29 + 32 + 35 + 38 + 41 + 44
Leya [2.2K]
The sum of the finite arithmetic series of <span> 26 + 29 + 32 + 35 + 38 + 41 + 44 is 245. Arithmetic series is a sequence of number such that the difference between any term and the previous term is a constant number. When we sum a finite number of terms in the arithmetic series, we get the finite arithmetic series. </span>
4 0
3 years ago
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An ellipse has a center at the origin, a vertex along the major axis at (0, 17), and a focus at (0, −8). Which equation represen
Marianna [84]
We know that
<span>Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide.
</span>Then 
a²<span> will go with the </span>y<span> part of the equation
</span>Also, since the focus is 8 <span>units below the center, then </span><span>c = 8
</span>since the vertex is 17<span> units above, then </span><span>a = 17
</span>The equation b²<span> = a</span>²<span> – c</span>²<span> gives me
</span>b²=17²-8²-----> b²=225
the equation is
y²/a²+x²/b²=1------> y²/289+x²/225=1

the answer is
y²/289+x²/225=1

see the attached figure

7 0
3 years ago
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Question is in picture. Please show your work, so I can see how you got the answer.
weqwewe [10]
It’s D! they have one intersect
5 0
3 years ago
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