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Answers
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Answer:
4.y=
x
2
+4x−7
[
x
=
2
x
1
Differentiating w.r.t. x, we get
dx
dy
=
2
x
2
+4x−7
1
⋅
dx
d
(x
2
+4x−7)
=
2
x
2
+4x−7
1
(
dx
d
x
3
+
dx
d
4x−
dx
d
7)
=
2
x
2
+4x−7
1
⋅(2x+1−0)
=
2
x
2
+4x−7
2(x+2)
=
x
2
+4x−7
(x+2)
6.⇒V=43π(d2)3=16πd3, where d is diameter of the sphere. So, rate of change of the volume of a sphere with respect to its diameter is πd22 cubic units.
7. Given that sin
−1
(1−x)−2sin
−1
x=
2
π
let x=siny
∴sin
−1
(1−siny)−2y=
2
π
⇒sin
−1
(1−siny)=
2
π
+2y⇒1−siny=sin(
2
π
+2y)
⇒1−siny=cos2y
⇒1−siny=1−2sin
2
y as cos2y=1−2sin
2
y
⇒2sin
2
y−siny=0
⇒2x
2
−x=0
⇒x(2x−1)=0
⇒x=0,2x−1=0
⇒x=0,2x=1
⇒x=0,x=
2
1
But x=
2
1
does not satisfy the given equation
∴x=0 is the solution of the given equation.
8.Let x=cotθ⟹cot
−1
x
Now, tan
−1
{
1+x
2
−x}=tan
−1
{
1+cot
2
θ
−cotθ}
=tan
−1
{
cosec
2
θ
−cotθ}
=tan
−1
{cosecθ−cotθ}
=tan
−1
{
sinθ
1
−
sinθ
cosθ
}
=tan
−1
{
sinθ
1−cosθ
}
=tan
−1
⎩
⎪
⎪
⎨
⎪
⎪
⎧
2sin
2
θ
cos
2
θ
2sin
2
2
θ
⎭
⎪
⎪
⎬
⎪
⎪
⎫
=tan
−1
⎩
⎪
⎪
⎨
⎪
⎪
⎧
cos
2
θ
sin
2
θ
⎭
⎪
⎪
⎬
⎪
⎪
⎫
=tan
−1
(tan
2
θ
)=
2
θ
=
2
1
cot
−1
x
Answer:
In parallelogram ABCD,
Let ∠A be x. Then, ∠B will be 2x−15
o
We know that, in a parallelogram sum of adjacent angles are supplementary.
∴ ∠A+∠B=180
o
⇒ x+2x−15
o
=180
o
⇒ 3x=195
o
⇒ x=65
o
.
∴ ∠A=65
o
⇒ ∠B=2x−15
o
=2×65
o
−15
o
=115
o
.
We know that, in parallelogram opposite angles are equal.
∴ ∠A=∠C and ∠B=∠D
∴ ∠C=65
o
and ∠D=115
o
.
∴ The measure of all angles of a parallelogram are 65
o
,115
o
,65
o
and 115
o
.
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