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Three days before yesterday is Wednesday.

Then, today is 4 days after Wednesday So, today is Monday.

Therefore, two days after tomorrow is 3 days after Monday which is Friday.

(3) is correct option

4 Is the best answer

3 days before yesterday is Wednesday,,that means 4days back was Wednesday. then today will be Sunday.

2 days after tomorrow will be 3 days. Then 3 days after Sunday will be wednesday.

answer is Wednesday,..!!

Wednesday is ryt ans

9^(x+2) -9^x = 240

=> 9^x(9²-1) = 240=> 9^x(80) = 240=> 9^x = 240/80 = 3=> 3^2x = 3=> 2x = 1=> x = 1/2

First find prime factors of 9696 = 2*2*2*2*2*3

sqroot (96) = 2*2*sqroot (2*3)= 4 sqroot (6)

How To Solve

How To Simplifying

I think the answer is false

the answer is 5 is rational no.

x²+2 is minimum when x = 0 , so it will make the fraction maximum

so, local maxima is at x = 0 and the value is y = 6/(0+2) = 3.

px*x + 3x + q = 0 x =-1,-2 are roots so output will be 0. p(-1)(-1) + 3(-1) + q = 0 p + q = 3 p(-2)(-2) + 3(-2) + q = 0 4p + q = 6 4p + q - p - q = 6 - 3 3p = 3 p = 1 q = 2 Hence,q-p=2-1=1

x2-3x-m(m+3)=0D=b2-4ac=〖(-3〗2)- 4×1×-m(m+3)=9 – 4 ×-m2-3m=9+〖4m〗2+12m=4m2+12m+9=〖(2m)〗2+2×(2m)×3+32=(2m+3)2√(D ) =2m+3 x=(-b±√(b2-4ac)) / 2a=(3+(2m+3))/(2×1)= (6+2m)/2= 6/2+2m/2 =3+mx= (-b±√(b2-4ac))/2a = (3-(2m+3))/(2×1) = (3-2m-3)/2 =-mANS : (-M,M+3)

a) putting value you will get -9 as 3xy gives -3 and the bracket term will be 3 so thoer product is -9

b and c also

b and c answers

1

2

3

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices.

SOLUTION:

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

1

polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

So, both are polynomials.

Total distance covered = 7*4 = 28 km

Total time taken = (7/10 + 7/20 + 7/30 + 7/60)

= 42+21+14+7/60

= 84/60

= 7/5 hours

average speed = Total distance/total time

= 28/(7/5)

= 4*5

= 20 kmph

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50.418km is the right answer

50.418km is the correct answer

50 km 418 m is the right answer.

good

Average speed = Total Distance / Total TimeTotal time = 11 km/hrTotal distance = 70 km

Average speed = 70/11 = 6(4/11) km /hr

(a) is correct option

since 1km =1000m so 7km =7×1000 m= 7000malso 21 km=21×1000 m= 21000m.

We know, 1 km= 1000 metres So,1) 7 km= (7*1000)metres = 7000 metres2) 21 km = (21*1000)metres = 21000 metres.

Since 1 Kilometre-1000mSo,7 KM will be 7000m and 21 KM will be 21000

1 kilometre=1000 metre7 kilometre=7000 metre

Thank you

1)The total number of students = 35 The number of girls = 10 The number of boys = 35-10 → 25

⇒ probability = total number boys/ total number of students ⇒ probability = 25/35 ⇒ 5/7 ∴ 5/7 is the probability of winning of boys.

Here we have to prove that the square has the greatest area for the same perimeter when a rectangle is considered.

So the perimeter P can be taken to be a constant P.

Use a variable x for the length. So the width is P/2 - x.

The area A = ( P/2 -x) *x.

Now to maximize the area, for a variable x, we have to find dA/dx.

dA/dx= P/2 - 2x.

Equate this to zero P/2 - 2x =0

=> x= P/4

So the length is P/4 and the width is P/2 - x = P/4.

Therefore for a given perimeter of a rectangle the area is the largest if the shape is that of a square with each side equal to the perimeter divided by 4.

OR

Letp be the perimeter , x be the length of the rectangle.

Then the width of the rectangle is (P-2x)/2.

Therefore the area of the rectangle with the perimeter p and the side x and (p-2x)/2 is given by:

A(x) = x*(p-2x)/2 = px/2 - x^2.

So by calculus, area A(x) is a function of perimeter p and A(x) is maximum, for x = c for which A'(c) = 0 and A"(c)< 0.

A'(x) = {px/2 - x^2}' = (px/2)' - (x^2)' .

So A'(x) = p/2 -2x. Equating to zero, we get p/2 - 2x = 0. Or

2x= p/2

x = p/4.

Again find A''(p/2) :

A'(x) = p/2-2x.

Differentiate:

A"(x) = 0-2 , which is negative for all x and so for x = p/4 also.

So A"(p/2) is negative for x= p/2 for which f'(p/2) = 0.

So, for a given perimeter p, x = p/4 gives the maximum area.

So , if p is the perimeter, the maximum area formed among the set of rectangles is the square with sides of length p/4 .

Steps of construction:1. Draw a line AB measuring 6.5 cm.2. Then draw a line through A taking 70° by protector. 3. Then by compass, cut a line measuring 8 cm placing the compass on both the points A and B. Then draw through A first.4. take a measurement of AB 6.5 cm. then through C draw an arc on point D.5. Then join the line BD.

BC=10m is the right answer

Steps of construction:1. Draw a line AB measuring 6.5 cm.2. Then draw a line through A taking 70° by protector. 3. Then by compass, cut a line measuring 8 cm placing the compass on both the points A and B. Then draw through A first.4. take a measurement of AB 6.5 cm. then through C draw an arc on point D.5. Then join the line BD.