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3 years ago

A butterfly starts at 20 m and changes -16m ?

Mathematics

1 answer:

miskamm [114]3 years ago

4 0

Answer:

it is at 4 meters

Step-by-step explanation:

20 - 16 = 4

Hope this helps!

Maybe brainliest?

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A line passes through (10,2) and (14,-22). Write the equation of the line in standard form.

jarptica [38.1K]

Answer:

The equation of the line in standard form is:

$6x + y = 62$

Step-by-step explanation:

Given the points

(10,2)

(14,-22)

Determining the slope between (10, 2) and (14, -22)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(10,\:2\right),\:\left(x_2,\:y_2\right)=\left(14,\:-22\right)$

$m=\frac{-22-2}{14-10}$

$m=-6$

The point-slope form of the line equation is

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

substituting the values m = -6 and the point (10, 2) in the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

$y - 2 = -6(x - 10)$

$y - 2 = -6x +60$

adding 2 to both sides

$y-2+2 = -6x + 60 + 2$

$y = -6x + 62$

We can write the equation in the standard form such as

Ax + By = C

Thus,

$y = -6x + 62$

adding -6x to both sides

$6x + y = -6x + 62 + -6x$

$6x + y = 62$

Therefore, the equation of the line in standard form is:

$6x + y = 62$

4 0

3 years ago

How do you solve this? (#11/#12)

zepelin [54]

$\text{If}\ x_1,\ \text{and}\ x_2\ \text{are zeros of a polynomial function, then the function}\\\text{has equation}:\\\\f(x)=a(x-x_1)(x-x_2).\\----------------------------\\\\11.\ \text{We have the zeros}\ 3\ \text{and}\ i:\\\\f(x)=(x-3)(x-i)=(x)(x)+(x)(-i)+(-3)(x)+(-3)(-i)\\\\=x^2-xi-3x+3i=\boxed{x^2-(3+i)x+3i}\\\\12.\ \text{We have the zeros}\ -2\ \text{and}\ 2i:\\\\f(x)=(x-(-2))(x-2i)=(x+2)(x-2i)\\\\=x^2-(2i)^2=x^2-2^2i^2=x^2-4(-1)=\boxed{x^2+4}\\\\\text{used}\\\\(a-b)(a+b)=a^2-b^2\\\\i=\sqrt{-1}\to i^2=-1$