First to answer gets brainleist! and 15-20 points ! ! i only need answers no step by step

Can someone help me with one more pls I’ll comment on yours

just olya [345]

Answer:

A:

The perimeter of the parking lot is 302 yards.

The area of the parking lot is 5,544 yards squared.

B: Perimeter

C: Area

Step-by-step explanation:

A:

To find the perimeter of the parking lot, we add all the sides up. 2L + 2W = Perimeter.

2(63) + 2 (88) = 302

To find the area, we multiply the length by the width. 63 × 88 = 5,544.

B: We need to know the amount of chain needed to surround the parking lot. In other words, we need to line the perimeter with a chain.

C: We would be paving the entire area of the parking lot.

Hope this helps,

❤Nicki❤

6 0

3 years ago

How do you solve for the central area of a polygon?

ycow [4]

Getting the central area of a polygon is not possible. Probably the question refers to the central angle. In a a regular polygon, getting the central angle is that you need to use the formula.

Central Angle of a polygon = 360 degrees / number of sides of a polygon.
Why 360 degrees? Because, a circle has a total of 360 degrees. And a circle is a polygon with no sides.

5 0

2 years ago

How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

2 years ago

Helppppppppppppppppppppppppppppp

iren [92.7K]

To start, the first thing you need to know is that when you add decimals, you must line the decimal up, meaning, if you were adding 2.7+4.1, you have to make sure 2 is lined up with 4, and 7 is lined up with 1 so that the decimals align. Then, you add straight down from right to left and you get 6.8 as your answer.

3 0

3 years ago

Read 2 more answers

Is this a function, explain answer

likoan [24]

No. There are two spots in the vertical line. (Vertical line test)

5 0

2 years ago

Read 2 more answers