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3 years ago

What is AB? a. 6cm b. 9cm c. 4cm d. 2.2cm

Mathematics

1 answer:

Fed [463]3 years ago

7 0

Answer:

AB = 4 cm

Step-by-step explanation:

The tangent and secant are drawn from an external point to the circle.

Then the square of the measure of the tangent is equal to the product of the external part of the secant and the entire secant.

let AB = x, then

x(x + 5) = 6²

x² + 5x = 36 ( subtract 36 from both sides )

x² + 5x - 36 = 0 ← in standard form

(x + 9)(x - 4) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 9 = 0 ⇒ x = - 9

x - 4 = 0 ⇒ x = 4

However x > 0 ⇒ x = 4 ⇒ b = 4 CM

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Fill the table using this function rule.

ElenaW [278]

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3 years ago

Y = 2x – 5 2y – 4x =

nikklg [1K]

Answer:

2y-4x

Step-by-step explanation:

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3 years ago

Read 2 more answers

the large Square above has area 9 is divided into 9 smaller squares of equal area what is the length of how strong and bold

Kay [80]

If one square is divided into 9 smaller equal squares, then they have to be arranged in 3 lines of 3, that is 3 smaller equal squares per side of the original big square. That said, the area of the big square is equal to the multiplication of 3 small squares sides times 3 small squares sides, call x the length of the small squares.
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area = 9 = 3x*3x
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8 0

3 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

3 years ago

what is the answer please help me

kenny6666 [7]

I can’t see the entire question, but i’m going to go out on a limb and say it’s divide, y-value, x-value

3 0

3 years ago