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3 years ago

When I was 6 my sister was half my age Now I'm 70 how old is my sister?

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

6 0

She would be 67...always going to be 3 years younger

joja [24]3 years ago

6 0

You were 6, she was 3. You are now 70, she is now 67. The answer is 67.

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sleet_krkn [62]

Answer:

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3 0

3 years ago

#7 Cecilia made a fruit basket for her grandmother. There were 3 bananas and 6 oranges in the basket. What is the ratio of the n

castortr0y [4]

The answer is A.

explanation:

the obvious ratio is 3:6, because there are 3 bananas and 6 oranges. if you simplify 3:6 it’s 1:2.

hope i helped :)

4 0

4 years ago

Read 2 more answers

Marianna [84]

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Step-by-step explanation:

Let us revise some logarithmic rules

$log(a)^{n}=nlog(a)$
$log(ab)=log(a)+log(b)$
$nlog(a)+mlog(b)=log[(a)^{n}][(b)^{m}]$

∵ 10 log(x) + 5 log(64)

- At first re-write 10 log(x)

∴ 10 log(x) = $log(x)^{10}$

- Then re-write 5 log(64)

∴ 5 log(64) = $log(64)^{5}$

∴ 10 log(x) + 5 log(64) = $log(x)^{10}$ + $log(64)^{5}$

- Use the 3rd rule above to make it single logarithm

∵ $log(x)^{10}$ + $log(64)^{5}$ = $log[(x)^{10}][(64)^{5}]$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(64)^{5}]$

∵ 64 = 2 × 2 × 2 × 2 × 2 × 2

∴ We can write 64 as $2^{6}$

∴ $(64)^{5}=(2^{6})^{5}$

- Multiply the two powers of 2

∴ $(64)^{5}=(2)^{30}$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(2)^{30}]$

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

6 0

3 years ago

Pls help will give brainly and please don't pick random

Step2247 [10]

Answer:

1.0738 x 10^-2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

How can one halfx − 5 = one thirdx + 6 be set up as a system of equations? 2y + x = −10 3y + x = 18 2y + 2x = −10 3y + 3x = 18 2

Doss [256]

Answer:

2y-x= -10

3y-x = 18

Step-by-step explanation:

The correct option is . 2y-x = -10 , 3y-x = 18

one half x − 5 = 1/2 (x-5)

one third x + 6 = 1/3 *(x+6)

1/2 x-5 = 1/3x+6 =y

y= 1/2x-5

y = 1/3x+6

Now,

y=x/2 -5 equation 1

y = x/3 +6 equation 2

By taking L.C.M of the first equation we get:

y=x/2 -5

y= x-10/2

Now multiply both terms by 2.

2y=x-10

2y-x= -10

Now lets solve second equation:

Take L.C.M of the second equation:

y = x/3 +6

y=x+18/3

Multiply both sides by 3

3y= x+18

3y-x = 18

Therefore the system of equations we get is:

2y-x= -10

3y-x = 18 ....

4 0

3 years ago