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3 years ago

14

The length of a rectangular patio is 14 feet more than its width, 2w. The area of a patio, A(w), can be represented by the funct

ion

1 answer:

nydimaria [60]3 years ago

5 0

Answer: 4w^2 + 28w.

Step-by-step explanation:

Width of the rectangle= 2w

Length if the rectangle= 2w+14

Area of a rectangle= length×width

= 2w(2w+14)

= 4w^2 + 28w

The area of the rectangular patio A(w) is 4w^2 + 28w.

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Find the value of x in a triangle . If the hypotenuse of the triangle is 10 , the opposite is x+2 and the adjacent is x.

Sauron [17]

Answer:

x = 6

Step-by-step explanation:

using Pythagoras' identity in the right triangle.

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , that is

x² + (x + 2)² = 10² ← expand parenthesis on left side and simplify

x² + x² + 4x + 4 = 100 ( subtract 100 from both sides )

2x² + 4x - 96 = 0 ( divide through by 2 )

x² + 2x - 48 = 0

(x + 8)(x - 6) = 0 ← in factored form

equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 6 = 0 ⇒ x = 6

however, x > 0 , then x = 6

8 0

2 years ago

What is the distance between the points (3, 7) and (15, 16) on a coordinate

AysviL [449]

answer: a 13 units. its 13 units bc 13+3=16 soooooooooooo yeahhhhhhhh

6 0

4 years ago

Please help me to prove this!

Ymorist [56]

Answer: see proof below

<u>Step-by-step explanation:</u>

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

<u>Proof LHS → Middle:</u>

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

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

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

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

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

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

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

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

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

LHS = Middle $\checkmark$

<u>Proof Middle → RHS:</u>

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

Middle = RHS $\checkmark$

3 0

3 years ago

Perimeter 94 cm and length 39. What’s the width

ruslelena [56]

Answer:

8

Step-by-step explanation:

Take the length and multiply it by 2. then take your answer and multiply it by the perimeter. Then, take that answer and divide it by 2. Hope this helps :)

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3 years ago

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6.6 x 1021 in standard notation

Gre4nikov [31]

Answer:

6,600,000,000,000,000,000,000

Step-by-step explanation:

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3 years ago