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valentinak56 [21]
3 years ago
13

The owners of an amusement park want to add more rides and an arcade in an open field.Beloe is a sketch of their plan. the owner

s do not know all of the measurements of the field

Mathematics
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

Step-by-step explanation:

Formula to get the area of a rectangle,

Area = Length × Width

Area of rides area = 30 × n

                              = 30n yds²

Area of the arcade = 30 × 40 yds²

Area of the whole field = Area of rides + Area of Arcade

                            Area  = [30n + 30(40)] square yards

Length of whole field = (n + 40) yds

Width of whole field = 30 yds

Area of the whole field = 30(n + 40) square yards

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What is the solution to the equation?
stiv31 [10]

Answer:

x = - 71

Step-by-step explanation:

x - 12 = - 83

+12          +12  (adding 12 to both sides)

x = - 71

7 0
3 years ago
When born, a baby weighed 8.5 pounds and gained 1.6 pounds per week for the first 10
ladessa [460]

Answer:

I believe the answer is C

Step-by-step explanation:

if the baby weighted 8.5 pounds when born then that is the initial value therefore the baby will be weighted 1.6 times 10 weeks or w

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5 0
3 years ago
Find all solutions of each equation on the interval 0 ≤ x < 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

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u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
3 years ago
Read 2 more answers
A rectangle has an AREA of 36in2​ and the height measures 9 in. What is the base measurement?
RSB [31]
Area =bh
36in2=b(9)
——— ——
9 9

36
—— =4
9

Base= 4
7 0
2 years ago
Can somebody please help me
Levart [38]

Answer:

462 sq in

Step-by-step explanation:

3((14 2/3)(10.5))

3(154)

462

7 0
3 years ago
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