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Scanner is the CORRECT ANSWER I THINK so

because printer , pointing device and omr is USED for output and scanner is INPUT

Your ANSWER is 783


hope it helps

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Uuuy 7986 6hhgg yfhj huytgjk vytthk GGG 8976

N=40
median of even no.=n/2+n/2+1
n/2 =40/2=20
median =20TH +21TH TERM.
the SUM of 20th and 21th term will be the median

16x+85= 17x+51
or, X= 34
so, NEW AVG. SCORE is 37

In the figure ABCD is a square and EF is parallel to BD YES BEIS ALSO parallel to DF because all the sides are equal in a square

this is CHEMISRTY not MATHS

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No they would not be ENOUGH as STUDENTS are 45 and PACKETS are only 12 with 20 beads left
Hope this HELPS you

A=(1,3,4)
B=(2,3)
REMAINING are(1,3) (3,2) (4,2),

1/2 and 3/4 can be WRITTEN as 50/100 and 75/100

so RATIONAL NUMBERS between are 51/100 , 52/100, 53/100, 54/100 , 55/100.....

I HOPE this HELPS...

3.476
hope this will HELP you

Please mark it BRAINLIEST I need one brainliest to go in NEXT level

Let's say the no of DAYS the worker works is60-x and the no of days a worker stays idle=x. as he is given 20 rupees EVERY day for WORKING and forfeits 3 rupees for STAYING idle. The required equation is (60-x)20-x(3)=280=>1200-20x-3x=280
=>1200-280=23x=>920=23x=>x=40 So he remains idle for 40 days and works for 20 days.

Given Q and R are the mid points of SIDES AB and BC of ∆ABC respectively

∴ QR||AC  [Mid-point THEOREM]

Given BE⊥AC

∴ BE⊥QR 

In ∆ABH and Q and P are the mid points of sides AB and AH respectively.

∴ QP||BH  [Mid-point theorem]

∴QP||BE

⇒QP⊥QR  (BE⊥QR)

⇒∠PQR = 90°

<P>The answer is
85/1000 (p and Q are INTEGERS and q≠0).

Hope it helps you.

X+6=5x-2=>4x=8=>x=2. So PERIMETER of a TRIANGLE is x+6+5x-2+7x+4=>13x+6-6=>13x=26cm is your perimeter .

0,1,1,2,3,4,5,6.......
these NUMBERS ocuur on the faily TREE of a HONEY bee.

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0.0000112378859916362761846462819

Let no of notebooks = x
so NUMBER of PENS = 25x
ATQ
25X×X=3600
25x²=3600
X²=3600÷25
x²=144
x=12

No of notebooks = 12
No of pens = 12×25 = 300

Hope it helps:)
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Given : 5ax + 6by = 28 ------ (1)

Given : 3ax + 4by = 18 ------ (2)

On SOLVING (1) * 3, & (2) * 5, we get

= > 15ax + 18by = 84

= > 15ax + 20by = 90

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

-2by = -6

y = 3/b.

Substitute y = 3/b in (1), we get

= > 5ax + 6by = 28

= > 5ax + 6b(3/b) = 28

= > 5ax + 18 = 28

= > 5ax = 28 - 18

= > 5ax = 10

= > ax = 2

= > x = (2/a).

Therefore, the value of x = (2/a) and y = (3/b).

Hope this helps!

There are 41 .
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we NEED to sell seventy-eight APPLES for 2340 to GAIN 30 PERCENT

The IDENTITY is
(a-b)^2 = a^2 - 2AB + b^2
(7X-9Y)^2
(7x)^2 - 2×7x×9y + (9y)^2
49x^2 - 126xy + 81y^2

Coordinate of mid point of the LINE segment JOINING A (3,6) is in quadrant 1
coordinate of mid point of the line segment joining B (-5,-10) is in quadrant 2

(a) A(2,-3)&B(5,6) in the ratio 2:3

x= (m*x2 + N*x1)/(m+n)

x=(2*5 + 3*2)/5

x=(10+6)/5

x=16/5

y= (m*y2 + n*y1)/(m+n)

y= (2*6 - 3*3)/5

y= (12-9)/5

y= 3/5

(x,y) = (16/5,3/5)

(b) A(-2,-3)&B(3,7) in the ratio 7:2

x= (m*x2 + n*x1)/(m+n)

x= (7*3 - 2*2)/9

x= 17/9

y= (m*y2 + n*y1)/(m+n)

y= (7*7 - 2*3)/9

y= 43/9

(x,y) = (17/9,43/9)

As FORMULA is c/5=f-32/9

Hi,
Here is your answer in the ATTACHMENT.......
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And the ANSWER is in PIC and answer is

I HOPE it is HELPFUL for you

tan^2 60 + COS^2 45

tan 60 = rt(3) ; cos 45 = 1/(rt(2))

so,

(rt(3))^2 + 1 / ((rt(2))^2)

= 3+(1/2)

(6+1)/2

7/2 = 3 1/2 or 3.5

Hello! ! !

Hey dear here is your answer ___________

➖Function is f(X) = 1/(1 - 2sinx)
function is DEFINED only when 1 - 2sinx ≠ 0


➖Let 1 - 2sinx = 0
sinx = 1/2 = sinπ/3


➖we know, if sinФ = sinα then, Ф = nπ + (+1)ⁿ α
so, x = nπ + (-1)ⁿπ/3 , where n is INTEGERS


➖Hence, domain of f(x) is all real NUMBER except {nπ + (-1)ⁿπ/3}


➖e.g., domain ∈ R - {nπ + (-1)ⁿπ/3 , n∈I}


____________________________

HOPE THIS ANSWER WILL HELP U...

@Neha...

The answer is
First we will MULTIPLY them with their total income

He SPEND on food is

32000*1/4=8000

He spend on house rent

32000*3/10=9600

He spend on education on children

32000 * 5/21=7619.04

Now we will add the product

8000+9600+7619.04=25219.04

Now we will SUBTRACT that from the total income

32000-25219.04=6780.96

The money LEFT with him is 6780.96.

Which CLASS BOOK is this QUESTION CONTAIN.

The above IMAGE CONTAINS the ANSWER for the QUESTION