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3 years ago

Uh yeah I need help can someone please explain to me please and thank you :)

Mathematics

2 answers:

Rzqust [24]3 years ago

6 0

Answer:

53/36

Step-by-step explanation:

56/36 is already in it's simplest form

Ivanshal [37]3 years ago

5 0

Answer:

1 17/36

Step-by-step explanation:

Turn each fraction into their LCD (least common denominator)

5/9 turns into 20/36

11/12 turns into 33/36

add 20 + 33 = 53

Keep denominator 53/36

53/36 simplified as mixed number = 1 17/36

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Which of the following conjectures is false?<br>Please give a counterexample!<br>Thank youuu

MakcuM [25]

Which of the following conjectures is false?

Solution: The false conjecture is:

J. The sum of two odd numbers is odd

Explanation:

F. The product of two even numbers is even.

For example: Let us take two even numbers 4 and 6.

The product of two even numbers is:

$4 \times 6 = 24$

Therefore, the product is also even number. Hence the conjecture is true.

G. The sum of two even numbers is even.

For example: Let us take two even numbers 4 and 6.

The sum of two even numbers is:

$4+6 = 10$

Therefore, the sum is also even number. Hence the conjecture is true.

H. The product of two odd numbers is odd.

For example: Let us take two odd numbers 3 and 5.

The product of two odd numbers is:

$3\times5 = 15$

Therefore, the product is also odd number. Hence the conjecture is true.

J. The sum of two odd numbers is odd.

For example: Let us take two odd numbers 3 and 5.

The sum of two odd numbers is:

$3+5= 8$

Therefore, the sum is not an odd number. Hence the conjecture is false.

6 0

3 years ago

Help friends please :(

olga nikolaevna [1]

The correct answer is: [B]: " 2⁶ * 2^(3/10) * 2^(8/100) " .
________________________________________________________

7 0

3 years ago

(05.01 LC)*PLEASE ANSWER 20 POINTS +BRANLIEST PLEASEEE!!!

Bess [88]

C) 1/25 for every 1 it goes up 25. therefore 1/25! hope I helped!!! brainliest would be much appreciated because I am trying to rank up!!!

3 0

3 years ago

Read 2 more answers

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$