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3 years ago

Please help, thanks I’m advance!

Mathematics

1 answer:

netineya [11]3 years ago

3 0

Answer:

Step-by-step explanation:

(x+2)©= x©+4x+4

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Say it with symbols a pool has lenght of 3s and a width of 2s. square tiles that are 1 ft. by 1ft. will be placed around the poo

Pachacha [2.7K]

Border of the pool means "perimeter".
P = 2L + 2W (L represents length, W represents width)
P = 2(3s) + 2(2s)
P = 6s + 4s
P = 10s

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3 years ago

TELL ME WHAT X IS PLEASE

Kaylis [27]

X=6 2+1=3 and 3 x 2=6 so x is 6

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4 years ago

5u^2 + 6 = -13u please help

Komok [63]

Answer:

u = -2, -3/5

General Formulas and Concepts:

Algebra I

Standard Form: ax² + bx + c = 0
Factoring
Finding roots

Step-by-step explanation:

Step 1: Define equation

5u² + 6 = -13u

Step 2: Solve for u

Rewrite: 5u² + 13u + 6 = 0
Factor: (u + 2)(5u + 3) = 0
Find roots: u = -2, -3/5

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3 years ago

The vertices of a pre-image are located at points (-3,3), (0,0), and (-2,-1). If the image triangle has vertices with coordinate

Dennis_Churaev [7]

It is an expansion by [y3, x3]

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4 years ago

Read 2 more answers

Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

Shape ABC

Assuming you want the area and perimeter of the shaded part of the shape only...

Area

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

Perimeter

Arc length = $r\theta$ (where r is the radius and $\theta$

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

Shape ABCD

Area

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

Perimeter

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago