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ASHA 777 [7]
3 years ago
8

8 times the sum of half a number and 12 is 30

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0
8(1/2n + 12) = 30 <== ur equation
4n + 96 = 30
4n = 30 - 96
4n = - 66
n = -66/4
n = - 16 1/2 or -16.5 <== ur solution
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6 more than 5 times a number y
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How do you simplify this expression? -12 divided by 3•(-8+(-4)^2-6)+2
stepan [7]

I good accumulator to use would be symbolab.com

The answer would be 3072/5629

5 0
3 years ago
CAN SOMEONE PLEASE HELP ME!!!!!
USPshnik [31]

Answer:

Step-by-step explanation:

2{5x²-15+(-9xy²)}-(2y²+4x-xy²)+3x²

=2{5x²-15-9xy²}-(2y²+4x-xy²)+3x²

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8 0
2 years ago
Two families went to the event. The first family bought two children's tickets and one adult ticket and paid a total of 8.2 euro
blagie [28]

Answer:

Step-by-step explanation:

step a system of two equations         c = child ticket     a = adult ticket

eq 1)  2c + 1a = 8.2            multiply by 2

eq 2) 3c + 2a = 14.1

I will multiply eq 1 times TWO and subtract eq 2 from eq 1a)

         eq 1a)  4c + 2a = 16.4

          eq 2)  3c + 2a = 14.1

subtract  (4c - 3c) + (2a -2a)  = 16.4 - 14.1  

                    c      +       0       =   2.3 euros     for one child ticket

Now find the adult ticket price,  plug 2.3 for c into eq 1)

     eq 1)  2c + 1a = 8.2

     eq 1)  2(2.3) + 1a = 8.2         solve for a

                 4.6  + a   =  8.2         substract 4.6 from both sides

                             a =  8.2 - 4.6

                               =   3.6 euros    for one adult ticket

double check using eq 2)    we know c and a values

    eq 2)  3c + 2a = 14.1

    eq 2)  3(2.3) + 2(3.6) = 14.1

                   6.9 + 7.2    =  14.1

                          14.1    =  14.1

5 0
3 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
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