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2 years ago

PLSSSS HELLPPPP MEEEEE I WILLL GIVEEE BRAINLIESSSTTT!!!!

Mathematics

2 answers:

earnstyle [38]2 years ago

6 0

Answer:

$\displaystyle m=3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (6, 8)

Point (5, 5)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [SF]: $\displaystyle m=\frac{5-8}{5-6}$
[Fraction] Subtract: $\displaystyle m=\frac{-3}{-1}$
[Fraction] Divide: $\displaystyle m=3$

Kamila [148]2 years ago

3 0

Answer:

Step-by-step explanation:

-3/-1 = 3

the slope of a line passing through (6,8) and (5,5) is 3.

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write the standard form of the equation of the circle for which the endpoints of a diameter are (0,0) and (4,-6)

Elenna [48]

Given:

The endpoints of a diameter are (0,0) and (4,-6).

To find:

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Solution:

The endpoints of a diameter are (0,0) and (4,-6). So, the length of the diameter is

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Now, radius is half of the diameter.

$r=\dfrac{2\sqrt{13}}{2}$

$r=\sqrt{13}$

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$Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Center=\left(\dfrac{0+4}{2},\dfrac{0+(-6)}{2}\right)$

$Center=\left(\dfrac{4}{2},\dfrac{-6}{2}\right)$

$Center=\left(2,-3\right)$

Standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

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The center of the circle is (2,-3) and radius is $\sqrt{13}$ . So,

$(x-2)^2+(y-(-3))^2=(\sqrt{13})^2$

$(x-2)^2+(y+3)^2=13$

Therefore, the standard form of the circle is $(x-2)^2+(y+3)^2=13$ .

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guapka [62]

Answer:

Step-by-step explanation:

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