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mr Goodwill [35]
2 years ago
5

Quick question. solve for x. show all work

Mathematics
2 answers:
kkurt [141]2 years ago
6 0
75.................................
Ipatiy [6.2K]2 years ago
6 0
75 Because someone said so
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34 &lt; 5x + 9 &lt; 44<br> Inequality
Ksivusya [100]

Answer:

5 < x < 7

Step-by-step explanation:

- Move all terms not containing  x  from the center section of the inequality.

25  <  5 x  <  35

- Divide each term in the inequality by  5 .

5 < 1x < 7

5 < x < 7

i hope this helps :)

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2 years ago
Help with parallel lines, please?
lianna [129]

No, in order for the lines to be parallel, the slopes have to be the same, it does not matter what the slopes are as long as they are the same slope. The lines could still be parallel if the slopes are negative.

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3 years ago
What is the range of the following relation?<br><br> {(-1, 1), (-2, 2), (-3, 3), (-4, 4), (-5, 5)}
jeka94
As the graph approach negative infinity the graph move to positive infinity as the graph approach positive infinity the graph move to negative infinity
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A line has a slope of -2 and passes through the point (1,-2). what is an equation of this line ?
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4 0
2 years ago
Read 2 more answers
Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (
kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2}) is given as continuous function, so there exist lim_{(x,y)\rightarrow(0,0)}f(x,y) and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})

we know that cos^2A+sin^2A=1, so we have that

f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)

lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19

So f(0,0)=ln19.

8 0
3 years ago
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