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Here, Height of parallelogram And triangles DAE, EDF , ECF and FCB remain same = h , As all these triangles and parallelogram lies between same parallel lines PQ | | RS .

Also given AE = EF = FB ,

SO,

AB = AE + EF + FB

AB = AE + AE + AE

AB = 3 AE .

We know area of parallelogram = Base Height So,Area of parallelogram ABCD = AB × h ---------- ( 1 )

AndWe know Area of triangle = 1/2×Base×Height

So,

Area of triangle DAE = 1/2×AE× h , Substitute value from AB = 3 AE , we get

Area of triangle DAE = 1/2×AB/3×h

Area of triangle DAE = 1/6×AB×h , Substitute value from equation 1 , we get

Area of triangle DAE = 1/6AreaofparallelogramABCD ( Hence proved )

Mozzarella needed = 3 1/4 ÷ 1/2

= 13/4 ÷ 1/2

= 13/4 * 2

= 13/2

= 6 1/2 cups

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Milk required to make 2/3 of recipe = 2/3(3/4) = 1/2 litres of milk

1/2 vinegar in her recipe made earlier

Total phone calls made by company = 798146

Number of free phone calls= 86197

Therefore,Number of phone calls company charged for = 798146 - 86197= 711952

2 l 400 ml = 2400 ml 1 glass of 300 ml so number of glasses in 2400 ml = 2400ml/300ml = 8

please give clear picture

a=2; b=3; a^2+b^2=(2)^2+(3)^2=4+9=13

0+(1/2)(1/2)+(root(3)/2)(root(3)/2)+1*1=0+1/4+3/4+1=1+1=2

(x+3)3= x3+ 27 + 9x(x+3) =x3+ 27 +9x2+27x.

Therefore coefficient of x is 27.

no

cos^-1(-1/2)=-2/1; sin^-1(1)=sin(1); cos^-1( cos2pie/3)+sin^-1(sin2 pie/3)=2/1+1/1=3/1=3

2π/3 + 2π/34π/3is correct answer

cos^-1(-1/2)=-2/1; sin^-1(1)=sin(1); cos^-1( coa2pie/3)+sin^-1(some pie/3)=2/1+1/1=3/1=23

1

2

Let y = sin^-1(1/2) sin y = 1/2sin y = sin (pi/6)

Range of principal value of sin^-1 is (-pi/2, pi/2)

Hence, the principal value is pi/6

Let cosec-¹(2) = ∅ where ∅ is the principal branch of cosec-¹(2) therefore cosec∅ = 2 since the range of principal value of branch of cosec-¹x is [+π/2,π/2] -{0} therefore -π/2≤∅≤π/2, ∅≠0

Now cosec∅ = 2 = cosecπ/6 thus ∅ = π/6 as cosecπ/6 = 2 and π/6 € [-π/2,π/2]-{0}

Radius= 1.4 cmarea= πr^2= 22/7*1.4*1.46.16cm^2

thankyou

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secx + tanx=p; 1+ sinx/ cosx=p; 1+cosx= pcosx; squaring both sides; 1+Sinx^2+2sinx=p^2 cosx^2; 1+sinx^2+2sinx=p^2(1-sinx^2); 1+ sinx^2+2 sinx=p^2-p^2sinx^2; sinx=-2+-V4+4(1-p^2)(1+p^2)/2(1-p^2)=-2+-2p^2/2(1+p^2)=-1+p^2/1+p= p^2-1/p^2+1

2/7+5/7=7/7=1one is right answer

-5/4+5/4=0right answer is zero

1 is the best answer

So Compound interest will be 35123.20 - 25000= Rs 10123.20

3........,............

1 is the right answer

Pen = 2x+3

Pencil = y-3

5 pens = 5(2x+3)

= 10x+15

3 pencils =3(y-3)

=3y - 9

Together they cost = pens + pencils

=10x + 15 + 3y - 9

= 10x + 3y + 15 - 9

= 10x + 3y +6

Given secA + tan A = pRecall the identity, sec^2A – tan^2A = 1⇒ (secA + tan A)(secA – tanA) = 1⇒ p × (secA – tanA) = 1∴ (secA – tanA) = 1/p

A straight line segment can be formed by joining any two points in space.

A 10 दिन लेता है। B 12 दिन लेता है। C 15 दिन लेता है। तीनो मिलकर करे तो 1/10+1/12+1/15 = (12+10+8)/120=30/120=1/4 4 दिन लेंगे।

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√1/cos^2theta+1/sin^2theeta√sin^2theta+cos^2theta/sin^2 theta. cos^2 theta1/costheta. sinthetasectheta. cosecthetaRHSsintheta/costheta+costheta/sinthetasin^2theta+cos^2theta/sin theta . cos theta1/sintheta. costhetaLHS=RHS

माध्य ज्ञात करने के लिए इनको पहले एक निश्चित करना व्यवस्थित करना होगा

19,25,30,31,32,35,48,51,59

32 माध्यक होगा

1. A straightline segmentcan be drawn joining any two points.

2. Any straightline segmentcan be extended indefinitely in a straightline.

3. Given any straightline segment, acirclecan be drawn having the segment asradiusand one endpoint as center.

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Sum of angles of a quadrilateral = 360=> 42 + 54 + x + x + 70 = 360=> 2x = 360 - 166=> 2x = 194=> x = 97 degrees.

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Postulate is a kind of axiom which is applied in a particular field.

postulate is a kind of axiom which is applied in a particular field .