Useful Log Rules:
- log(A*B) = log(A)+log(B) .......... log rule 1
- log(A/B) = log(A) - log(B) .......... log rule 2
- log(A^B) = B*log(A) ................... log rule 3
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Part (a)
All logs shown below are base 3.
log(500) = log(5*100)
log(500) = log(5*10^2)
log(500) = log(5)+log(10^2) .... use log rule 1
log(500) = log(5) + 2*log(10) .... use log rule 3
log(500) = 1.4650 + 2*2.096 ....... substitution
log(500) = 5.657
<h3>Answer: 5.657</h3>
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Part (b)
All logs shown below are base 3.
log(2) = log(10/5)
log(2) = log(10) - log(5) .... use log rule 2
log(2) = 2.096 - 1.4650 ....... substitution
log(2) = 0.631
<h3>Answer: 0.631</h3>
Answer: 94.3 meters
Step-by-step explanation:
Answer:
-24
Step-by-step explanation:
P=26/27
Q=-2/3
10q-18p
10(-2/3)-18(26/27)=-24
Answer: For the sum of 130
First: $90
Second: $40
Step-by-step explanation:
We write equations for each part of this situation.
<u>The Total Charge</u>
Together they charged 1550. This means 1550 is made up of the first mechanics rate for 15 hours and the second's rate for 5 hours. Lets call the first's rate a, so he charges 15a. The second's let's call b. He charges 5b. We add them together 15a+5b=1550.
<u>The Sum of the Rates</u>
Since the first's rate is a and the second is b, we can write a+b=130 since their sum is 130.
We solve for a and b by substituting one equation into another. Solve for the variable. Then substitute the value into the equation to find the other variable.
For a+b=130, rearrange to b=130-a and substitute into 15a+5b=1550.
15a + 5 (130-a)=1550
15a+650-5a=1550
10a+650-650=1550-650
10a=900
a=$90 was charged by the first mechanic.
We substitute to find the second mechanic's rate.
90+b=130
90-90+b=130-90
b= $40 was charged by the second mechanic
Answer:
the area of the triangle is 20 units, and the answer to the expression is also 20, so its equal to each other. 20=20
the area of the trapezoid is 38.5 units
Step-by-step explanation:
