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4 years ago

A data set consists of the following data points:

1 answer:

8 0

Since the slope of the line is already given, by plotting the three points and drawing a best fit line using the given slope and extending the line to hit the x-axis, the y-intercept can be determined. Line-fitting tools can also be used; common scientific calculators have that feature.
So, the answer is D. 16

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Identify the type of slope and then find the slope of the line.

Irina18 [472]

Answer:

-1/2, negative slope

Step-by-step explanation:

As per graph it is visible that the slope is negative, it goes down when x-value increases

Use two points to find the slope:

(-4, -1) and (0, -3)

Slope formula:

m = (y2 - y1)/(x2 - x1)

Value of the slope:

m = (-3 - (-1))/(0 - (-4)) = -2/4 = -1/2

4 0

3 years ago

Match the number with its opposite. Match Term Definition 9.2 A) 2.9 −2.9 B) −4.1 −1.4 C) −9.2 4.1 D) 1.4

jenyasd209 [6]

We have these opposite pairs

9.2 and -9.2
2.9 and -2.9
1.4 and -1.4
4.1 and -4.1

So all we're doing is matching each positive number with its negative version. In terms of a visual, the opposite of a number is mirrored over 0 on the number line. So for instance, the opposite of 2 is -2, with each being two units away from 0 on the number line.

8 0

3 years ago

What type of angle is R?

tester [92]

Answer:

obtuse

Step-by-step explanation:

its over 90 degrees

8 0

3 years ago

Read 2 more answers

HURRY 15(-74a+51)=15(7a+456) a.6 b.-5 c.-2 d.5

Airida [17]

Answer:

-5 because of I'm right I am always right

5 0

3 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

4 years ago