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emmasim [6.3K]
3 years ago
7

Three hundred cars drove over a bridge in 23 minutes. At that rate, how

Mathematics
2 answers:
anzhelika [568]3 years ago
7 0

Answer:

1800

Step-by-step explanation:

300 cars in 23 minutes so 300/25 = how many cars each minute

that equals 13.04347....

so on your calculator multiply that by 138 which gives you 1800 cars

Paraphin [41]3 years ago
3 0

Answer:

138/23×300=1800cars drove over the bridge

Step-by-step explanation:

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Answer:

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Step-by-step explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m  ×  b^n  =  b^(m + n) ⇒ in multiplication if they have same base

  we add  the power

2- b^m  ÷  b^n =  b^(m – n) ⇒  in division if they have same base we

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3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

   them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same  

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5- b^(-m) = 1/(b^m)  (for all nonzero real numbers b) ⇒ If we have

   negative power we reciprocal the base to get positive power

6- If  a^m  =  a^n  ,  then  m  =  n ⇒ equal bases get equal powers

7- If  a^m  =  b^m  ,  then  a  =  b    or    m  =  0

# Logarithmic rules:

1- log_{a}b=n-----a^{n}=b

2- loga_{1}=0---log_{a}a=1---ln(e)=1

3- log_{a}q+log_{a}p=log_{a}qp

4- log_{a}q-log_{a}p=log_{a}\frac{q}{p}

5- log_{a}q^{n}=nlog_{a}q

* Now lets solve the problems

# 3^{x+1}=9^{x+3}

- Change the base 9 to 3²

∴ 9^{x+3}=3^{2(x+3)}=3^{2x+6}

∴ 3^{x+1}=3^{2x+6}

- Same bases have equal powers

∴ x + 1 = 2x + 6 ⇒ subtract x and 6 from both sides

∴ 1 - 6 = 2x - x

∴ -5 = x

* The solution x = -5

# ㏒(9x - 2) = ㏒(4x + 3)

- If ㏒(a) = ㏒(b), then a = b

∴ 9x - 2 = 4x + 3 ⇒ subtract 4x from both sides and add 2 to both sides

∴ 5x = 5 ⇒ divide both sides by 5

∴ x = 1

* The solution is x = 1

# log_{6}(5x+4)=2

- Use the 1st rule in the logarithmic equation

∴ 6² = 5x + 4

∴ 36 = 5x + 4 ⇒ subtract 4 from both sides

∴ 32 = 5x ⇒ divide both sides by 5

∴ 6.4 = x

* The solution is x = 6.4

# log_{2}x+log_{2}(x-3)=2

- Use the rule 3 in the logarithmic equation

∴ log_{2}x(x-3)=2

- Use the 1st rule in the logarithmic equation

∴ 2² = x(x - 3) ⇒ simplify

∴ 4 = x² - 3x ⇒ subtract 4 from both sides

∴ x² - 3x - 4 = 0 ⇒ factorize it into two brackets

∴ (x - 4)(x + 1) = 0 ⇒ equate each bract by 0

∴ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

OR

∵ x + 1 = 0 ⇒ subtract 1 from both sides

∴ x = -1

- We will reject this answer because when we substitute the value

 of x in the given equation we will find log_{2}(-1) and this

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* The solution is x = 4

# log_{4}11.2=x

- You can use the calculator directly to find x

∴ x = 1.7427

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∴ e^{8x}=4.6

- Insert ln for both sides

∴ lne^{8x}=ln(4.6)

- Use the rule ln(e^{n})=nln(e) ⇒ ln(e) = 1

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The amount c in grams of a 100 gram sample of carbon-14 remaining after t years is represented by the equation c= 100 (0.99988)t
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Answer: The amount of carbon-14 remaining after 4 years is 99.95 grams.

Step-by-step explanation:

Hi, to answer this question we simply have to substitute t=4 in the equation given and solve for c.

c= 100 (0.99988)^t

c =100 (0.99988)^4

c = 100 x 0.999520086

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The amount of carbon-14 remaining after 4 years is 99.95 grams.

Feel free to ask for more if needed or if you did not understand something.

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