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3 years ago

Which is bigger? 1/2 or 7/15?

Mathematics

1 answer:

kogti [31]3 years ago

3 0

Answer:

$\frac{1}{2} \: > \: \frac{7}{15}$

Step-by-step explanation:

We want to determine which of the following fractions is bigger.

The fractions are:

$\frac{1}{2} \: and \: \frac{7}{15}$

We collect LCM to get;

$\frac{15}{30} \: and \: \frac{14}{30}$

Now that the denominators are the same we can easily compare.

$\frac{15}{30} \: > \: \frac{14}{30}$

This means that:

$\frac{1}{2} \: > \: \frac{7}{15}$

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Rewrite 4 1/2=2 <br> please

Luda [366]

The logarithmic expression of 4^(1/2) = 2 is $\log_4(2) = \frac 12$

<h3>How to rewrite the expression?</h3>

The expression is given as:

4^(1/2) = 2

Take the logarithm of both sides

log(4^(1/2)) = log(2)

Apply the change of base rule

1/2log(4) = log(2)

Divide both sides by log(4)

1/2 = log(2)/log(4)

Change the base

$\frac 12 =\log_4(2)$

Rewrite as:

$\log_4(2) = \frac 12$

Hence, the logarithmic expression of 4^(1/2) = 2 is $\log_4(2) = \frac 12$

Read more about logarithmic expression at:

brainly.com/question/24211708

#SPJ1

8 0

2 years ago

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

2 years ago

True or false and if false what would be the correct term be?

vagabundo [1.1K]

It’s True.. you take the “extreme” variable from each proportion and cross multiply them before setting them equal to the product of the two “means” (which are just the other 2 numbers in the proportions).

3 0

3 years ago

PLEASE HELP ASAP! I have reposted this 2 times and no one will help me.:( please answer all 4 thank you so much!! i need this do

erik [133]

Answer:

5. DE

6. 3 cm

7. 20 in

8. ΔDCF

5 0

3 years ago

A weather service records 28.8 inches of rain over a 12 month period. What is the average rainfall per month for the year?

Basile [38]

Answer:

B) 2.4 inches

Step-by-step explanation:

The total amount of rain is 28.8 inches over 12 months. To find how much rain falls averagely in <em>one </em>month you need to divide the total by the amount of months. 28.8/12 = 2.4

6 0

2 years ago