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4 years ago

What is the surface area of the prism in square inches?

Mathematics

2 answers:

astra-53 [7]4 years ago

8 0

Answer:

4 inches deep, 5 inches high and 15 inches across

Harrizon [31]4 years ago

6 0

Answer:

To find the surface area of a prism, the problem can be approached in one of two ways.

1. Through an equation that uses lateral area

2. Through finding the area of each side and taking the sum of all the faces

Using the second method, it's helpful to realize rectangular prisms contain 6 faces. With that, it's helpful to understand that there are 3 pairs of sides. That is, there are two faces with the same dimensions. Therefore, we really only have three sides for which we need to calculate areas:

Faces 1 & 2:

15⋅8=120

Faces 3 & 4:

6⋅8=48

Faces 5 & 6:

15⋅6=90

Now, we can add up the areas of all six sides:

120+120+48+48+90+90=516

The surface area is 516units2.

Step-by-step explanation:

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Find the area of the triangle with vertices: q(3,-4,-5), r(4,-1,-4), s(3,-5,-6).

maw [93]

Answer:

(√6)/2 square units

Step-by-step explanation:

The area of a triangle is half the magnitude of the cross product of the vectors representing adjacent sides.

QR = (4-3, -1-(-4), -4-(-5)) = (1, 3, 1)

QS = (3 -3, -5-(-4), -6-(-5)) = (0, -1, -1)

The cross product is the determinant ...

$\text{det}\left|\begin{array}{ccc}i&j&k\\1&3&1\\0&-1&-1\end{array}\right|=-2i+j-k$

The magnitude of this is ...

|QR × QS| = √((-2)² +1² +(-1)²) = √6

The area of the triangle is half this value:

Area = (1/2)√6 . . . . square units

3 0

3 years ago

Solve. Simplify your answer. 5/8 x 2/3 = ?

Finger [1]

10/24 simplify to 5/12

5 0

3 years ago

A city’s bus line is used more as the urban population density increases. The more people in an area, the more likely bus lines

nevsk [136]

Answer:

(A) The residents of Belmont are more likely to use public transportation because the city has the highest population density.

Step-by-step explanation:

correct on edge

7 0

3 years ago

Please help me to prove this!

Ymorist [56]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ B + C = π - A

→ C + A = π - B

→ C = π - (B + C)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → Middle:

$\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)$

$\text{Sum/Difference:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)$

$\text{Double Angle:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)$

$\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]$

$\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]$

$\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)$

$\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)$

LHS = Middle $\checkmark$

Proof Middle → RHS:

$\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)$

Middle = RHS $\checkmark$

3 0

3 years ago

Find the measure of the incanted angle to the nearest degree

tiny-mole [99]

Answer:

34⁰

Step-by-step explanation:

let unknown angle be x

cos x=19/23

cos x=0.826

x=cos inverse of 0.826

x=34.3⁰

4 0

3 years ago