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2 years ago

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The function g is given in three equivalent forms. Which form most quickly reveals the zeros (or "roots") of the function? Choos

e 1 answer: (Choice A) A g(x)=-2(x+4)^2+18 (Choice B) B g(x)=-2x^2-16x-14(Choice C) C g(x)=-2(x+1)(x+7)

1 answer:

koban [17]2 years ago

8 0

Answer:

C. $g(x) = -2 \cdot (x + 1)\cdot (x+7)$

Step-by-step explanation:

The polynomial form that reveals most quickly the zeroes is the form of a product of binomials. That is:

$g(x) = \Pi\limits_{i=1}^{n} (x-r_{i})$

Where $\Pi$ is the product function and $r_{i}$ is the i-th root of the polynomial.

Hence, $g(x) = -2 \cdot (x + 1)\cdot (x+7)$ resembles a form that is close to the form described above. The right option is C.

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2 years ago

Mathematics, what's the answer?

sleet_krkn [62]

Answer:

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Step-by-step explanation:

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2 years ago

Find the value of x

Paladinen [302]

$\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star$

x = 12°

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

$\qquad❖ \: \sf \:2x + 6 + 5x + 90 = 180$

[ by linear pair ]

$\qquad❖ \: \sf \:7x + 6 = 180 - 90$

$\qquad❖ \: \sf \:7x + 6 = 90$

$\qquad❖ \: \sf \:7x = 90 - 6$

$\qquad❖ \: \sf \:7x = 84$

$\qquad❖ \: \sf \:x = 84 \div 7$

$\qquad❖ \: \sf \:x = 12$

$\qquad \large \sf {Conclusion} :$

$\qquad❖ \: \sf \:x = 12 \degree$

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1 year 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

Use your knowledge on the triangle proportionality theorem to find the missing side length indicated. Be sure to explain the pro

ikadub [295]

Answer:

? = 4.29

Step-by-step explanation:

Remark

Let x be the question mark. You get the proportionality by using the dimensions of the little triangle to the dimensions of the large triangle.

Equation

14/(14+ 4) = 15/(x + 15) Combine like terms on the left

14/18 = 15/(x + 15) Cross multiply

14*(x + 15) = 18 * 15 Simplify the right

14(x + 15) = 270 Divide by 14

x + 15 = 270 / 14

x + 15 = 19.29 Subtract 15 from both sides

x = 19.29 - 15

x = 4.29

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2 years ago