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3 years ago

Someone I need help this is my last time on this app I need help (correct answers only )

Mathematics

2 answers:

Airida [17]3 years ago

8 0

Answer:

FDDU

Step-by-step explanation:

RUR

Harrizon [31]3 years ago

3 0

Answer:

ok options?

Step-by-step explanation:

You might be interested in

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,

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

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

$\to x$ )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence $a_n$ such that:

$a_n\ belongs\ to\ Q$

and

$a_n\to \alpha$

$f(a_n)=ka_n$

( since $a_n$ belongs to Q )

Let f is continuous at x=α

This means that:

$f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha$

This means that:

$f(\alpha)=k\alpha$

This means that:

f(x)=kx for every x∈ R

4 0

3 years ago

a bag contains 5 red balls and some Blue Balls if the probability of drawing a blue ball is double that of red ball find the num

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

Let the number of blue balls be x.

Number of red balls = 5

Total number of balls = x + 5

P (getting a red ball) = 5/x+5

P (getting a blue ball) = x/x+5

Given that

2(5/x+5) = x/x+5

10(x+5) = x^(2) +5x

x^(2)-5x-50=0

x(x-10)(x+5)

Now either x-10 = 0 or x+5= 0

But number of balls cannot be negative.

Hence, the number of blue balls is 10.

5 0

4 years ago

Consider the quadratic equation. x^2=4x-5. How many solutions does the equation have?

Vesna [10]

x^2=4x-5
subtract 4x from both sides
x^2-4x=-5
add 5 to both sides
x^2-4x+5=0

input into quadratic formula which is x= $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ or $\frac{-b- \sqrt{b^2-4ac} }{2a}$

si ax^2+bx+c
so a=1
b=-4
c=5
input
$\frac{-(-4)+ \sqrt{-4^2-4(1)(5)} }{2(1)}$ = $\frac{4+ \sqrt{16-20} }{2(1)}$ = $\frac{4+ \sqrt{-4} }{2}$ = $\frac{4+ \sqrt{4} times \sqrt{-1} }{2}$ $\frac{4+2 times \sqrt{-1} }{2}= \frac{6 times \sqrt{-1} }{2}=3 times \sqrt{-1} [\tex][\tex]\sqrt{-1}$ representeds by 'i' so solution is 3i

then if other way around then wyou would do
$\frac{-(-4)- \sqrt{-4^2-4(1)(5)} }{2(1)}$ = $\frac{4- \sqrt{16-20} }{2(1)}= \frac{4- \sqrt{-4} }{2} =\frac{4- \sqrt{4} times \sqrt{-1} }{2}= \frac{4-2 times \sqrt{-1} }{2}=\frac{2 \sqrt{-1} }{2}= \sqrt{-1}$ and [\tex]\sqrt{-1} [/tex] is represented by i

the solution is x=3i or i (i= $\sqrt{-1}$ )
but i is not real, it is imaginary so there are no real solution so the answer is C

3 0

4 years ago

On a bird-watching trip, Mei and Jackie counted 68 birds.

defon

Answer:

The answer is 70% if you round it.

Step-by-step explanation:

45 out of 68 is 66 but if you're rounding it the answer is 70%.

5 0

3 years ago

What is the standard form of (7-5i) (2+3i)

ladessa [460]

Answer:

(7-5i)( 2+3i)

=7(2)+ 7(3i)-5i(2) -(5i)(3i)

= 14 +21i -10i +15

= 29 +11i

3 0

4 years ago