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malfutka [58]
3 years ago
13

Do you guys know how to solve with equation?

Mathematics
2 answers:
AlexFokin [52]3 years ago
5 0

Answer:

{-8, 3/4}

Step-by-step explanation:

One by one, set the factors equal to zero and solve for v:

8 + v = 0 yields v = -8, and

4v - 3 = 0 yields 4v = 3 and v = 3/4

BigorU [14]3 years ago
3 0
Wouldn’t it be 4v^2+29v-24, like the picture showed.

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A baker uses 5.5lbs of flour daily. How many ounces of flour will he use in two weeks.
atroni [7]
8 oz in 1 poount
therefor, multiply number of poounds by 8 to get ounces
5.5 times 8=44
this is in 1 day

how much in 2 weeks
1 week=7 days
2 weeks=7 times 2=14

1 day=44oz
14 day=44oz times 14=616oz

the answer is 616oz
4 0
3 years ago
What is the net monthly cash flow
Eddi Din [679]

Answer:

what?


Step-by-step explanation:


7 0
3 years ago
What is a decimal that is equivalent to the fraction 44/100​
balandron [24]

Answer:

\frac{44}{100 }  \\  = 0.44

=> A decimal that is equivalent to the fraction 44/100 is <u>0</u><u>.</u><u>4</u><u>4</u>

4 0
3 years ago
Read 2 more answers
Is this correct???? TYSM
vfiekz [6]

Yep, this is all correct!

-6/2 = y/4

y = (-6×4)/2

y = -24/2

y = -12


x was multiplied by 2

and y was also multiplied by 2

5 0
4 years ago
A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and
RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

  • A_t = A_0e^r^t
  • where Aₜ = mass at time t, A₀ = mass at time 0,  r = decay constant (rate), t = time  

Our known variables are:

  • 1998 to the year 2004 is a total of t = 6 years.
  • The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

  • 270=360e^-^r^(^6^)
  • 270=360e^-^6^r
  • \frac{3}{4} =e^-^6^r
  • \text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)
  • \text{ln} (\frac{3}{4} )=-6r
  • r=-\frac{\text{ln}\frac{3}{4} }{6}
  • r=0.04794701

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

  • A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)
  • A_t=270e^-^0^.^3^3^5^6^2^9^0^7
  • A_t=193.01982213

The expected mass of the sample in the year 2011 would be 193 mg.

3 0
3 years ago
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