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2 years ago

For the data shown in the scatter plot below, identify a point apart from the main cluster?

Mathematics

1 answer:

Kazeer [188]2 years ago

7 0

Answer:

C. (8, 7)

Step-by-step explanation:

<u>There are 2 points that are apart from the main cluster:</u>

(8, 7) and (8.5, 7.5)

<u>One of them is included in the answer choices:</u>

C. (8, 7)

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Find the distance and midpoint for (2, 0, -2) and (5, -4, 6)

sweet-ann [11.9K]

Hi, It is well simples:

The distance between two points is :

d( A, B) = √[ (xb - xa)^2+(yb - ya)^2+(zb - za)^2]

Then,

A = (xa, ya, za )

B = (xb , yb , zb)

We know:

A = ( 2 ,0 ,-2)

And,

B = (5 , -4 , 6)
___________

xb - xa = 5-2 <=> 3

yb - ya = -4 - 0 <=> -4

zb - za = 6 - (-2) <=> 8

Then us stay:

d( A, B) = √[ (3)^2 + (-4)^2+(8)^2]

= √( 9 + 16 + 64)

= √( 25 + 64)

= √(89)

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2 years ago

Choose 3 values that would make this inequality true. 9 - n ≥ 4

Alenkasestr [34]

Answer: Could be 0,1,2,3,4,5

Step-by-step explanation: 9-n has to be greater or equal to four. 9 minus 4 is five. So all of the answers must be positive numbers that are less than five.

6 0

2 years ago

Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

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1 year ago

What is the value of cd in the equation 32cd=11cd-42?

USPshnik [31]

Answer:

32cd-11cd=-42

21cd=-42

cd=-42/21

cd=(-2)

the answer is d

7 0

2 years ago

Hey how do you get from standard form to vertex form?

vlabodo [156]

Explanation:

Conversion of a quadratic equation from standard form to vertex form is done by completing the square method.

Assume the quadratic equation to be $\mathbf{ax^{2}+bx+c=0}$ where x is the variable.

Completing the square method is as follows:

send the constant term to other side of equal $\mathbf{ax^{2}+bx=-c}$
divide the whole equation be coefficient of $\mathbf{x^{2}}$ , this will give $\mathbf{x^{2}+\frac{b}{a}x=- \frac{c}{a}}$
add $\mathbf{(\frac{b}{2a})^{2}}$ to both side of equality $\mathbf{x^{2}+2\times\frac{b}{2a}x+\frac{b}{2a}^{2}=-\frac{c}{a}+\frac{b}{2a}^{2}}$
Make one fraction on the right side and compress the expression on the left side $\mathbf{(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}}$
rearrange the terms will give the vertex form of standard quadratic equation $\mathbf{a(x+\frac{b}{2a})^{2}-\frac{b^{2}-4ac}{4a}=0}$

Follow the above procedure will give the vertex form.

(NOTE : you must know that $\mathbf{(x+a)^{2}=x^{2}+2ax+a^{2}}$ . Use this equation in transforming the equation from step 3 to step 4)

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2 years ago