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Vladimir79 [104]
2 years ago
6

5. Change the following mixed numbers to improper

Mathematics
1 answer:
bonufazy [111]2 years ago
3 0
A.23/6
b.23/8
c.26/5
d.64/9
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Find the product of 543.1187 and 100.
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A 54,311.87
It moves the decimal places to the left by two.
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3 years ago
write 5 to the 8 power as a quotient of two exponential terms with the same base in four different ways. use negative and/or zer
shusha [124]

Answer:

jhfjdhsuhufhre

Step-by-step explanation:

a) 3x+2(41-5x)=33            as y=41-5x is given

3x+82-10x=33

-7x+82=33

-7x=33-82

-7x=-49        cut the negative signs as its LHS and RHS

x=49/7=7

b)-5x+3(3x-3)=3   as y=3x-3 is given

-5x+9x-9=3

4x-9=3

4x=3+9

x=12/4=3

c)4x+11-5x=9

-x+11=9

-x=9-11

-x=-2         cut the negative signs as its LHS and RHS

x=2

d) 5y+ -5-2y=-11      as x=-5-2y is given

3y+-5=-11

3y=-11+5

3y=-6

y=-6/3=-2

e)-5x-4(2x+1)=35       as y=2x+1

-5x-8x+4=35

3x+4=35

3x=35-4

3x=31

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x=10.3

7 0
3 years ago
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has i
bija089 [108]

Answer: \bold{a)\ P(t)=P_o\cdot e^{t\cdot ln(2.7)}}

               b) 5314

               c) ln 2.7

               d) 4.6 hrs

<u>Step-by-step explanation:</u>

P(t) = P_o\cdot e^{kt}\\\\\bullet \text{P(t) is the number of bacteria after t hours} \\\bullet P_o\text{ is the initial number of bacteria}\\\bullet \text{k is the rate of growth}\\\bullet \text{t is the time (in hours)}\\\\\\270=100\cdot e^{k(1)}\\2.7=e^k\\ln\ 2.7=\ln e^k\\\boxed{ln\ 2.7=k}\\\\\text{So the equation to find the number of bacteria is: }\boxed{P(t)=P_o\cdot e^{t\cdot ln(2.7)}}\\\\\\P(4)=100\cdot e^{4\cdot ln(2.7)}\\.\qquad =\boxed{5314}

10,000=100\cdot e^{t\cdot ln(2.7)}\\100=e^{t\cdot ln(2.7)}\\ln\ 100=ln\ e^{t\cdot ln(2.7)}\\ln\ 100=t\cdot ln(2.7)\\\dfrac{ln\ 100}{ln\ 2.7}=t\\\\\boxed{4.6=t}

4 0
3 years ago
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