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Dimas [21]
3 years ago
6

I just need to see how to work this problem out, so I don't have to ask a question like this again..and I know what I'm doing ne

xt time! Thanks :)
20 pts!

−6x+12>24=
Mathematics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer by Mimiwhatsup:

-6x+12>24\\\mathrm{Subtract\:}12\mathrm{\:from\:both\:sides}\\-6x+12-12>24-12\\Simplify\\-6x>12\\Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)\\\left(-6x\right)\left(-1\right)

=-\frac{12}{6}\\\mathrm{Divide\:the\:numbers:}\:\frac{12}{6}=2\\=-2\\Answer is: x

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Who ever answers first gets branliest :))
scoundrel [369]
The answer is 36.86

(I think)
7 0
2 years ago
I WILL GIVE A CROWN JUST NEED HELP ASAP
pav-90 [236]

Answer:

$720

Step-by-step explanation:

First tutor = 30 x 10 = $300

Second tutor = 30 x 14 = $420

Total = 420 + 300 = $720

3 0
3 years ago
I REALLY NEED HELP ASAP!!!
kkurt [141]

m<P would be:

25

but m<PQC I forgot how to do... sorry!

8 0
3 years ago
what is the y-intercept of the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15
ohaa [14]

The slope of the given line is the coefficient of x, 3/5. The slope of the perpendicular line will be the negative reciprocal of that: -1/(3/5) = -5/3.

The point-slope form of the equation for a line of slope m through point (h, k) can be written ...

... y = m(x -h) +k

For your point and the slope found above, this becomes

... y = (-5/3)(x -15) -5

When x=0, this is

... y = (-5/3)(-15) -5 = 20

The y-intercept is 20.

3 0
3 years ago
Please Help !!!
solong [7]
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)

moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g


Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05

N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
5 0
3 years ago
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