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Marina CMI [18]

3 years ago

11

I just cheated on a test in a website called assessprep. I don't know if it could detect cheating and how it does that. (I'm wor

ried cuz my teacher told everyone to be honest on this)

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

i think itll be okay as long as you didnt copy it word for word

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Find the area of the shaded region. Round to the nearest tenth

Sophie [7]

Answer :

57770

Solution :

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3 years ago

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Triangle ABC is formed by the vertices A(1,2,-1), B(-3,-6,2)and C(3,-2,0).

Vesnalui [34]

Answer:

Point D: $(0,-4, 1 )$

<em>d</em> = √41

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Pre-Calculus</u>

Midpoint Formula [3D]: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} )$
Distance Formula [3D]: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Step-by-step explanation:

<u>Step 1: Define</u>

Point A(1, 2, -1)

Point B(-3, -6, 2)

Point C(3, -2, 0)

<u>Step 2: Find Point D</u>

Simply plug in your coordinates B and C into the midpoint formula to find midpoint

Substitute [MF]: $(\frac{-3+3}{2},\frac{-6-2}{2}, \frac{2+0}{2} )$
Add/Subtract: $(\frac{0}{2},\frac{-8}{2}, \frac{2}{2} )$
Divide: $(0,-4, 1 )$

<u>Step 3: Find distance </u><em><u>d</u></em>

Simply plug in the 2 coordinates A and D into the distance formula to find distance <em>d</em>

Substitute [DF]: $d = \sqrt{(0-1)^2+(-4-2)^2+(1+1)^2}$
Subtract/Add: $d = \sqrt{(-1)^2+(-6)^2+(2)^2}$
Exponents: $d = \sqrt{1+36+4}$
Add: $d = \sqrt{41}$

5 0

3 years ago

M+5n=7 how to solve equation

aivan3 [116]

M + 5n = 7
subtract 5n from each side

m = -5n + 7

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4 years ago

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jekas [21]

Answer:

slope: 1/3; y-intercept: 0

Step-by-step explanation:

using the equation y = mx + b we can find the slope and y-intercept

the slope is m and the y-intercept is b

in this case m is 1/3

there is no b so the y-intercept is 0

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