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zlopas [31]
3 years ago
10

Help on this plzz real quick

Mathematics
1 answer:
konstantin123 [22]3 years ago
7 0
★ LOGARITHMIC REDUCTIONS ★

{e}^{3x + 6}  = 8 \\ applying \:  log_{e} \: on \: both \: sides \\ 3x  + 6= ln8 \\ 3x =  \frac{ln8 - 6}{1}  \\ x =  \frac{ln8 - 6}{3}
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Complete the following proof Given: is the midpoint of , is the midpoint of Prove: =2
Vesnalui [34]

Answer:

Step-by-step explanation:

                        Statements                                    Reasons

1). M is the midpoint of segment AB        1). Given

   B is the midpoint of segment MD      

2). AM = MB and MB = BD                        2). Definition of midpoint

3). MD = MB + BD                                      3). Segment Addition Postulate

4). MD = MB + MB                                     4). Substitution property of of Equality

5). MD = 2MB                                            5). Simplify

Therefore, if M is the midpoint of segment AB, B is the midpoint of MD then MD = 2MB

5 0
3 years ago
Your promotional budget for the year is $300 for personal selling, $300 for sales promotion, \$1,500 for advertising, and $500 f
aleksandr82 [10.1K]

Answer:

to get the total promotion budget for the year we will add $300 for personal selling plus $300 for sales promotion plus $1500 for advertising plus $500 for image promotion to the a total of 2600 dollars

8 0
3 years ago
Melissa ate 1/2 of her salad for lunch and ate 4/9 of he salad for dinner. How much of her salad did she eat?
Alborosie

Answer:

13/18

Step-by-step explanation:

Since Melissa ate 1/2 of her salad during lunch, 1/2 of her salad should be left.

If she ate 4/9 of that 1/2 left over from lunch for dinner, 1/2 × 4/9 = 4/18

So she ate 1/2 + 4/18 of her salad in total.

9/18 + 4/18=

13/18 of her salad

7 0
3 years ago
250cm3 of fresh water of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the
natka813 [3]

Answer:

m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg

And we can do a similar procedure for the sea water:

m_{sea}= \rho_{sea} V_{sea}

And after convert the volume to m^3 we got:

m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg

And then the density for the mixture would be given by:

\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}

And replacing we got:

\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}

Step-by-step explanation:

For this case we can begin calculating the mass for each type of water:

m_{fresh}= \rho_{fresh} V_{fresh}

And after convert the volume to m^3 we got:

m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg

And we can do a similar procedure for the sea water:

m_{sea}= \rho_{sea} V_{sea}

And after convert the volume to m^3 we got:

m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg

And then the density for the mixture would be given by:

\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}

And replacing we got:

\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}

6 0
3 years ago
F= [-2 0] [0 8] [2 0] C= [12 0 3/2] [1 -6 7 ] what is fc?
Crank
Add them up i think i never tried this
7 0
3 years ago
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