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2 years ago

A. y= -9/8x-9/11 B. y= 9/8x + 11/8 C. y= -8x - 11 D. y = 8/9x - 11/9

Mathematics

1 answer:

Nutka1998 [239]2 years ago

6 0

Answer:

$y=\frac{8}{9}x- \frac{11}{9}$

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Sally was walking to Ocean Front Housing Complex, In miles, how far is her distance?

Vanyuwa [196]

Answer:

Sally is 10 miles

The coordinate ofthe deli is (1, -1)

Step-by-step explanation:

✍️Coordinate of Sally = (-3, 2)

Coordinate of Ocean Front = (5, -4)

The distance of Sally to Ocean Front Housing Complex = the distance between point (-3, 2) and (5, -4).

Use the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , to calculate the distance between (-3, 2) and (5, -4):

$d = \sqrt{(5 -(-3))^2 + (-4 - 2)^2}$

$d = \sqrt{(8)^2 + (-6)^2}$

$d = \sqrt{64 + 36}$

$d = \sqrt{100} = 10$

✅The distance of Sally is 10 miles

✍️the coordinate of Deli would be half the distance between Sally and Ocean Front.

Therefore, use the midpoint formula to calculate the coordinate of the midpoint between Sally (-3, 2) and Ocean Front (5, -4).

Midpoint of (-3, 2) and (5, -4).

$M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Plug in the values

$M(\frac{-3 + 5}{2}, \frac{2 +(-4)}{2})$

$M(\frac{2}{2}, \frac{-2}{2})$

$M(1, -1)$

✅The coordinate ofthe deli is (1, -1)

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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 $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

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2 years ago

Cherise has a bag of marbles for her little sister. It contains 10 orange marbles, 7 yellow marbles, and 3 green marbles. She dr

NemiM [27]

The probability that all marbles are yellow is
(7/20) (7/20) (7/20) = 343/8000

We are given
x = 2 and 3
u = 2.5
s = 0.5

Getting the z score
z = (2 - 2.5) / 0.5 and (3 - 2.5)/ 0.5
z = -1 and 1
The percentage is
1 - 0.1557 = 81.5%

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2 years ago

What is the solution to (x+6)(x+2)=60

Citrus2011 [14]

Distribute to eliminate the parenthesis

x^2 + 2x + 6x + 12 = 60

Combine like terms

x^2 + 8x + 12 = 60

Subtract 60 from both sides

x^2 + 8x - 48 = 0

Factor the equation

(x+12)(x-4)=0
x+12=0
x+4=0

x= -12 and -4

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3 years ago

I need help with this?? Anyone?

sergey [27]

- Question -

Which statement about solving inequalities is true?

- Answer -

A)

Adding the same value to both sides of an inequality does not change the solution set.

- The Wolf -

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2 years ago

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