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3 years ago

A graph that has a finite or limited number of data points is a

Mathematics

2 answers:

Neko [114]3 years ago

6 0

A qualitative graph because all the others have no ending

olchik [2.2K]3 years ago

3 0

Answer:

It is A. a discrete graph

Step-by-step explanation:

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(X-3)(x+2)=0 find the zeros

Ivanshal [37]

Answer: Min = (0.5, −6.25)

Step-by-step explanation: Standard form:

x2 − x − 6 = 0

Factorization:

(x + 2)(x − 3) = 0

Solutions based on factorization:

x + 2 = 0 ⇒ x1 = −2

x − 3 = 0 ⇒ x2 = 3

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4 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

ad-work [718]

Answer:

Well Where are the graphs

Step-by-step explanation:

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3 years ago

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12+4(8x-1) please help

MakcuM [25]

Answer:

32x+8

Step-by-step explanation:

You’re welcome :)

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3 years ago

Find the difference quotient of the exponential function, f(x) = e^x

Orlov [11]

Answer:

Step-by-step explanation:

The difference quotient in general looks like

$\frac{f(x+h)-f(x)}{h}$ Sometimes the symbol "delta x" $\Delta x$ is used instead of h.

For the exponential function, the difference quotient is

$\frac{e^{x+h}-e^x}{h}$

5 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago