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sp2606 [1]
2 years ago
6

Expand a logarithmic expression.

Mathematics
1 answer:
Korolek [52]2 years ago
4 0

we are given

log(z^3\sqrt{x^5y} )

step-1: Firstly , we will expand log terms by using property

log(a*b)=log(a)+log(b)

log(z^3)+log(\sqrt{x^5y} )

step-2: Now, we can bring exponent in front

log(a^n)=n*log(a)

=log(z^3)+log(x^5y)^{\frac{1}{2}}

=3log(z)+\frac{1}{2}log(x^5y)

step-3: we can again expand second term

=3log(z)+\frac{1}{2}(log(x^5)+log(y))

=3log(z)+\frac{1}{2}(5log(x)+log(y))

so, we will get

=3log(z)+\frac{1}{2}(5log(x)+log(y))...............Answer

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In ΔABC, BC = 12 and tan B = 0.58333. What is the length of the hypotenuse, to the nearest tenth?
choli [55]
Is there a pic showing ABC? it will be evil if BC is the hypotenuse itself lol

assuming dats not the case here,

tan B = CA/BC

so 0.58333 = CA/12

CA = <span>69.9996

the hypotenuse, AB = sqrt (12^2 + </span><span>69.9996^2)

= 71.02

= 71.0

</span>
6 0
3 years ago
What is the product? x^2-16/2x+8 multiplied by x^3-2x^2+x/x^2+3x-4
Contact [7]

Answer:

(x (x - 4) (x - 1))/(2 (x + 4))

Step-by-step explanation:

Simplify the following:

((x^2 - 16) (x^3 - 2 x^2 + x))/((2 x + 8) (x^2 + 3 x - 4))

The factors of -4 that sum to 3 are 4 and -1. So, x^2 + 3 x - 4 = (x + 4) (x - 1):

((x^2 - 16) (x^3 - 2 x^2 + x))/((x + 4) (x - 1) (2 x + 8))

Factor 2 out of 2 x + 8:

((x^2 - 16) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

x^2 - 16 = x^2 - 4^2:

((x^2 - 4^2) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

Factor the difference of two squares. x^2 - 4^2 = (x - 4) (x + 4):

((x - 4) (x + 4) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

Factor x out of x^3 - 2 x^2 + x:

(x (x^2 - 2 x + 1) (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

The factors of 1 that sum to -2 are -1 and -1. So, x^2 - 2 x + 1 = (x - 1) (x - 1):

(x (x - 1) (x - 1) (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

(x - 1) (x - 1) = (x - 1)^2:

(x (x - 1)^2 (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

((x - 4) (x + 4) x (x - 1)^2)/(2 (x + 4) (x + 4) (x - 1)) = (x + 4)/(x + 4)×((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)) = ((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)):

(x (x - 4) (x - 1)^2)/(2 (x + 4) (x - 1))

Cancel terms. ((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)) = ((x - 4) x (x - 1)^(2 - 1))/(2 (x + 4)):

(x (x - 4) (x - 1)^(2 - 1))/(2 (x + 4))

2 - 1 = 1:

Answer: (x (x - 4) (x - 1))/(2 (x + 4))

5 0
2 years ago
Read 2 more answers
An athletic league does drug testing of its athletes, 10 percent of whom use drugs. This test, however, is only 95 percent relia
Vera_Pavlovna [14]

Answer:

0.6786

0.3214

1 / 172

171 / 172

Step-by-step explanation:

Note: The tree diagram is attached below.

a) The athlete is a drug user, given that the test is positive P(D/+)

Using conditional Probability:

P ( D / + ) = P ( D and + ) / P ( +)

= (0.1 * 0.95) / (0.1 * 0.95 + 0.9*0.05)

= 0.6786

Answer: P ( D / + ) = 0.6786

b) The athlete is not a drug user, given that the test is positive P(D' / + )

Using conditional Probability:

P ( D' / + ) = P ( D' and + ) / P (+)

= (0.9 * 0.05) / (0.1 * 0.95 + 0.9*0.05)

= 0.3214

Answer: P ( D' / + ) = 0.3214

c) The athlete is a drug user, given that the test is negative P (D / - )

Using conditional Probability:

P ( D / - ) = P ( D and - ) / P (-)

= (0.1 * 0.05) / (0.9 * 0.95 + 0.1*0.05)

= 1 / 172

Answer: P ( D / - ) = 1 / 172

(d) The athlete is not a drug user, given that the test is negative P (D' / - )

Using conditional Probability:

P ( D' / - ) = P ( D' and - ) / P (-)

= (0.9 * 0.95) / (0.9 * 0.95 + 0.1*0.05)

= 171 / 172

Answer: P ( D / - ) = 171 / 172

4 0
2 years ago
What are examples of the distributive property?
Elena L [17]
Here is an example:

3(4x +7)
2( 3 +5) =2( 3 ) = 2 ( 5 )
8 0
3 years ago
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
kirza4 [7]

Answer:

Step-by-step explanation:Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

6 0
3 years ago
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