Can u pls help me with this question

What does AB congruent symbol CD mean? AB and CD here have lines over them How is it different than AB=CD?

masha68 [24]

Answer:

Suppose that we have two line segments, AB and CD. We know that they have the same length.

I know that AB¯¯¯¯¯¯¯¯=CD¯¯¯¯¯¯¯¯ means AB is identical to CD (aka. They are the same lines), and also that AB¯¯¯¯¯¯¯¯≅CD¯¯¯¯¯¯¯¯ means that AB and CD have the same size, but what does AB=CD mean?

Step-by-step explanation:

4 0

3 years ago

Six hundred kilograms of tea were packed in 500 g packets. how many packets of tea were there?

Shtirlitz [24]

Answer:

1200 packets of tea

Step-by-step explanation:

Its because theres 1000 grams in 1kg so it would be 600000 grams and you would divided 600000 over 500 and get 1200

Can you mark brainlyist?

5 0

2 years ago

Read 2 more answers

EFV and EDW are straight lines.<br>Find the value of x and of y.

Makovka662 [10]

Answers:

x = 72

y = 83

============================================================

Explanation:

Angle VFG is 50 degrees. The angle adjacent to this is angle EFG which is 180-50 = 130 degrees.

Angle HDW is 77 degrees. The supplementary angle adjacent to this is 180-77 = 103 degrees which is angle EDH.

Pentagon EFGHD has the following five interior angles

E = x
F = 130
G = 170
H = 65
D = 103

Note that angles F = 130 and D = 103 were angles EFG and EDH we calculated earlier.

For any pentagon, the interior angles always add to 180(n-2) = 180(5-2) = 180*3 = 540 degrees.

This means,

E+F+G+H+D = 540

x+130+170+65+103 = 540

x+468 = 540

x = 72

---------------------------------------

Now focus your attention on triangle THS

We see that the interior angles are

T = y
H = 65
S = 32

The angle H is 65 degrees because it's paired with the other 65 degree angle shown. They are vertical angles.

For any triangle, the angles always add to 180

T+H+S = 180

y+65+32 = 180

y+97 = 180

y = 180-97

y = 83

7 0

3 years ago

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

It’s MATH who knows how to do this i’m down bad

AysviL [449]

Answer/Step-by-step explanation:

1. 7x + 2 = 5x + 22 (alternate interior angles are congruent)

Collect like terms

7x - 5x = -2 + 22

2x = 20

2x/2 = 20/2

x = 10

2. 13x - 6 = 10x + 24 (alternate interior angles are congruent)

Collect like terms

13x - 10x = 6 + 24

3x = 30

3x/3 = 30/3

x = 10

3. (12x + 26)° + 46° = 180° (same side interior angles are supplementary)

12x + 26 + 46 = 180

12x + 72 = 180

12x + 72 - 72 = 180 - 72

12x = 108

12x/12 = 108/12

x = 9

4. (5x + 5)° + 135° = 180° (same side interior angles are supplementary)

5x + 5 + 135 = 180

5x + 140 = 180

5x + 140 - 140 = 180 - 140

5x = 40

5x/5 = 40/5

x = 8

4 0

3 years ago

Can u pls help me with this question ​

Can u pls help me with this question