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abruzzese [7]
3 years ago
6

Solve for x 3x + 2 y;y=

Mathematics
2 answers:
Nataly [62]3 years ago
5 0

Answer:

I beleive its 6

Step-by-step explanation:

Andreas93 [3]3 years ago
5 0

Answer:

3x +2= y; y=5. \\3x +2 = 5\\3x =5 -2\\3x = 3\\Divide-both-sides-of-the-equation-by 3\\\frac{3x}{3} =\frac{3}{3} \\x = 1

Step-by-step explanation:

x = 1

Step 1 : Substitute 5 for y in 3x +2= y

Step 2 : Collect like terms

Step 3 : Simplify

Divide both sides of the equation by 3

x = 1

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Answer:

on the 24th day they both will make chocolate the LCM of 8 and 12 is 24

4 0
3 years ago
Read 2 more answers
What is the volume of the cylinder?
jarptica [38.1K]

Answer:

v= 141.3

v= pi × r^2 × h

v= 3.14 × 3^2 × 5

v= 3.14 × 9 × 5

v=141.3

6 0
3 years ago
(EASY) I WILL MARK THE BRAINLYEST !!!
nydimaria [60]

Answer:

-8>x>6

Step-by-step explanation:

I 5x+5I +22> 57subtract 22 on both sides

I5x+5I >35

                                    SPLIT

5x+5>35                                              5x+5< -35 ( flip sign and 35 negative)

5x>30                                                   5x< -40

x>6                                                         x< -8

  Draw a number line Write  zero in the middle

From zero count  to the  left until- 8  make an open circle because- 8  is not included  From -8 draw  an arrow to the left (because x  is less than -8)  

From zero  count to the right until  6  Make an open circle bc 6     is not included.  From 6 draw an arrow to the right   (because x is more than 6)

PART 2 is not already solved?  

If you need to solve fot t

Divide both parts by pr

t= (l/ pr)                                      

4 0
3 years ago
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
Can someone please help me
Daniel [21]

Answer:

(-7.5, 3.5)

Step-by-step explanation:

midpoint \:  = ( \frac{x1 + x2}{2} \:  , \:  \frac{y1 + y2}{2} )

= ( \frac{ ( - 5) + ( - 10)}{2} , \frac{(5) + (2)}{2} )

= ( \frac{ - 15}{2} , \frac{7}{2} )

= ( - 7.5,3.5)

8 0
3 years ago
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