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NARA [144]
4 years ago
15

Use logarithmic properties and the fact that ln(2) ≈ 0.69 and ln(3) ≈ 1.10 to approximate the value of each of the following log

arithmic expressions. Do not use a calculator.
ln(6)
Mathematics
1 answer:
bija089 [108]4 years ago
3 0

Answer:

ln6 = 1.79

Step-by-step explanation:

The logarithm here has base e so it is a natural logarithm.

log_ex=lnx

lnx=y\\\Rightarrow e^y=x

According to the product rule of logarithms

ln(a\times b)=ln(a)+ln(b)

6=3\times 2

ln6=ln(3\times 2)\\\Rightarrow ln6=ln3+ln2\\\Rightarrow ln6=1.10+0.69\\\Rightarrow ln6=1.79

∴ ln6 = 1.79

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Answer:

The answer is A.

Step-by-step explanation:

1/2 (2x+6)>4

1x+3>4

1x>1

x>1

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What kinds of symmetry does the capital letter E have
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It has reflectional symmetry where the line of reflection (mirror line) is horizontal and it cuts through the middle of the letter
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The slope of EF¯¯¯¯¯ is −52.
9966 [12]

Based on the above, the segments that are perpendicular to EF are LM and NP.

<h3>Why is the segment are LM and NP perpendicular  to EF ?</h3>

Note that when two lines are perpendicular, we can say that;

M1 * M2 = -1 As M1 and M2 are known to be the slopes of the lines.

Therefore, when the the slope of EF is said to be −5/2, then  one can say that the slope of  the segment that is said to be  perpendicular to EF will have to be equal to m1*m2=-1, m2=-1/m1, m2=-1/(-5/2) or m2=2/5.

Scenario one:

JK , if J is at (3, −2) and K is at (5, −7)

To find the slope JK, then

m=(y2-y1)/x2-x1)

m=(-7+2)/(5-3)

m=-5/2

-5/2 is not equal to 2/5

Therefore,  JK is not perpendicular to EF

Scenario 2

Find GH , when G is at (6, 7) and H is at (4, 12)

To find the slope GH

m=(y2-y1)/x2-x1)

m=(12-7)/(4-6)

m=5/-2

m=-5/2

Since -5/2 is not equal to 2/5 then GH is not perpendicular to EF

Scenario 3:

Find LM , If L is at (1, 9) and M is at (6, 11)

To find the slope LM, then

m=(y2-y1)/x2-x1)

m=(11-9)/(6-1)

m=2/5

Since 2/5 is equal to 2/5

Then LM is perpendicular to EF

Scenario 4:

Find NP , if N is at (−3, 4) and P is at (−8, 2)

To find the slope NP, then

m=(y2-y1)/x2-x1)

m=(2-4)/(-8+3)

m=-2/-5

m=2/5

Since 2/5 is equal to 2/5.

Therefore,  NP is perpendicular to EF

Based on the above calculations, the segments that are perpendicular to EF are LM and NP.

See correct format of question written below

The slope of EF is −5/2 .

Which segments are perpendicular to EF?

Select all the right answers please

1. JK , where J is at (3, −2) and K is at (5, −7)

2. GH , where G is at (6, 7) and H is at (4, 12)

3. LM , where L is at (1, 9) and M is at (6, 11)

4. NP , where N is at (−3, 4) and P is at (−8, 2)

Learn more about segments from

brainly.com/question/10565562

#SPJ1

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2 years ago
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Answer:

D) 3^{6}

Step-by-step explanation:

if you multiply exponents, you actually add them

so... 4+2=6

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