All categories

3 years ago

3. approximate value of log 3 base 7

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

3 0

Answer:

0.56457

Step-by-step explanation:

log 3 with base 7 = (Log 3)/ (Log 7) = 0.47712125 / 0.845098040014

(Log 3) / (Log 7) = 0.56457

You might be interested in

Consider r (x) = StartFraction a x Superscript b Baseline + 8 Over c x Superscript d Baseline EndFraction, where a, b, c, and d

madam [21]

Answer:

The correct answer is 0 or choice A on edge

Step-by-step explanation:

right on edge

4 0

2 years ago

Find the percent cost of the total spent on each equipment $36, fees $158, transportation $59

maksim [4K]

Answer: Therefore, Option 'A' is correct.

Step-By-Step explanation:

Since we have given that

Amount spent on equipment = $36

Amount spent on fees = $158

Amount spent on transportation = $59

Total amount spent by them is given by

$36+158+59=\$253$

Percent cost of equipment of the total amount is given by

$\frac{36}{253}\times 100=14.2\%$

Percent cost of fees of the total amount is given by

$\frac{158}{253}\times 100\\\\=62.45\%$

Percent cost of transportation of the total amount is given by

$\frac{59}{100}\times 100\\\\=23.3\%$

Hence, Percent cost of the total spent on each is

14%, 62%, 23%.

Therefore, Option 'A' is correct.

6 0

2 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers

Which is larger .89 or .9

storchak [24]

.9 because the tenth place is 9 and the tenth place for the other one is 8 so its gonna be
.89 < .90 = .9

7 0

2 years ago

Read 2 more answers

Help with mathhh tt

Zepler [3.9K]

Answer:

wsp son

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

3. approximate value of log 3 base 7​

3. approximate value of log 3 base 7