Answer:
B x = sqrt(41)
Step-by-step explanation:
Since the large triangle is isosceles, the smaller triangles on the right side and left side are congruent right triangles.
The base of each smaller right triangle is 4 units long (8/2 = 4).
The vertical leg is 5.
Use the Pythagorean theorem to find x.
a^2 + b^2 = c^2
4^2 + 5^2 = x^2
16 + 25 = x^2
x^2 = 41
x = sqrt(41)
Answer: B x = sqrt(41)
Answer:
Step-by-step explanation:
Lateral surface area of the triangular prism = Perimeter of the triangular base × Height
By applying Pythagoras theorem in ΔABC,
AC² = AB² + BC²
(34)² = (16)² + BC²
BC = 
= 
= 30 in.
Perimeter of the triangular base = AB + BC + AC
= 16 + 30 + 34
= 80 in
Lateral surface area = 80 × 22
= 1760 in²
Total Surface area = Lateral surface area + 2(Surface area of the triangular base)
Surface area of the triangular base = 
= 
= 240 in²
Total surface area = 1760 + 2(240)
= 1760 + 480
= 2240 in²
Volume = Area of the triangular base × Height
= 240 × 20
= 4800 in³
Answer:
23.7 -2.5 X 8 = 23.7 - 20
= 3.7
Answer:
a) ∝A ∈ W
so by subspace, W is subspace of 3 × 3 matrix
b) therefore Basis of W is
={
}
Step-by-step explanation:
Given the data in the question;
W = { A| Air Skew symmetric matrix}
= {A | A = -A^T }
A ; O⁻ = -O⁻^T O⁻ : Zero mstrix
O⁻ ∈ W
now let A, B ∈ W
A = -A^T B = -B^T
(A+B)^T = A^T + B^T
= -A - B
- ( A + B )
⇒ A + B = -( A + B)^T
∴ A + B ∈ W.
∝ ∈ | R
(∝.A)^T = ∝A^T
= ∝( -A)
= -( ∝A)
(∝A) = -( ∝A)^T
∴ ∝A ∈ W
so by subspace, W is subspace of 3 × 3 matrix
A ∈ W
A = -AT
A = ![\left[\begin{array}{ccc}o&a&b\\-a&o&c\\-b&-c&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Do%26a%26b%5C%5C-a%26o%26c%5C%5C-b%26-c%260%5Cend%7Barray%7D%5Cright%5D)
=
![+c\left[\begin{array}{ccc}0&0&0\\0&0&1\\0&-1&0\end{array}\right]](https://tex.z-dn.net/?f=%2Bc%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%260%5C%5C0%260%261%5C%5C0%26-1%260%5Cend%7Barray%7D%5Cright%5D)
therefore Basis of W is
={
}
Answer:
4082
Step-by-step explanation:
Given
The composite object
Required
The volume
The object is a mix of a cone and a hemisphere
Such that:
<u>Cone</u>
---- radius (r = 20/2)

<u>Hemisphere</u>

The volume of the cone is:



The volume of the hemisphere is:



So, the volume of the object is:





