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zimovet [89]
2 years ago
13

( x + 3 )( y + 2 ) = xy - 5 ( x + 2 )( y - 3 ) = xy

Mathematics
1 answer:
Paraphin [41]2 years ago
4 0

Answer:

xy+2x+3y+6=xy-5 .............(1)

xy-3x+2y-6=xy.....,.,(2)

2x+3y=-11

-4x+2y=6

first equation multiply by 2

(2)*(2x+3y)=-11

4x+6y=-22

-4x+2y=6

8y=-16 y=(-2),,,,, x=(-2,5)

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Which of the values for x and y make the equation 2x + 3y + 4 = 15 true?
hjlf
X=1, y=3 because 2x1 is 2 and 3x3 is 9, 2 plus 9 equals 11 and 11 plus 4 equals 15
5 0
2 years ago
Read 2 more answers
What type of variable is the number of robberies reported in your city? Continuous attribute quantitative qualitative?
erastova [34]

solution:

Attribute is not type of variable, instead, attributes are the categories of a categorical variable. For example: if variable is gender, attributes are male , female.

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The number of robberies is quantitative because the value is numeric (discrete)

It is not qualitative because it is not nominal.  


4 0
3 years ago
Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).
andre [41]
Check the picture below.  So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\

\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}

7 0
3 years ago
Solve and check.
GarryVolchara [31]
R=104.

104/6=17.33333333333333
52/3=17.33333333333333

Hope this helped☺☺
8 0
3 years ago
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The sides of one right triangle are 6, 8 and 10. The sides of another right triangle are 10, 24 and 26. Determine if the triangl
Arlecino [84]
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5 0
3 years ago
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