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3 years ago

1. Given F ( x ) = 2/3 x - 4

2 answers:

7 0

What is the slope of this function?

answer
2/3
-----------------------
Given F ( x ) = 2/3 x - 42. What is the zero of this function?2/3 x - 4 = 0
2/3x = 4x = 6answer6
--------

3.Given F ( x ) = -2/3x - 4Is the point (-3, -6) a solution?
-2/3x - 4 = -2/3(-3) - 4 = 2 - 4 = -2
so when x = -3 y = - 2so answer is NO

KATRIN_1 [288]3 years ago

5 0

It's 2/3
-----------------------
Given F ( x ) = 2/3 x - 42. What is the zero of this function?2/3 x - 4 = 0
2/3x = 4x = 6answer6
--------

3.Given F ( x ) = -2/3x - 4Is the point (-3, -6) a solution?
-2/3x - 4 = -2/3(-3) - 4 = 2 - 4 = -2
so when x = -3 y = - 2 so answer is NO
-Hope It Helps! :3

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Find the angle between u = (-2,-5) and v = (5,2)

OleMash [197]

The angle between u = (-2,-5) and v = (5,2) is 134 degrees approximately.

Solution:

Given, two vectors are u = (-2, -5) and v = (5, 2)

We have to find the angle between two vectors.

We know that,

$a. b=\|a\| .\|b\| \times \cos \theta$

where $\theta$ is angle between vectors a and b

$\text { Now vectors are }(-2 \vec{\imath}-5 \vec{\jmath}) \text { and }(5 \vec{\imath}+2 \vec{\jmath})$

$(-2 \vec{\imath}-5 \vec{\jmath}) \cdot(5 \vec{\imath}+2 \vec{\jmath})=\sqrt{(-2)^{2}+(-5)^{2}} \times \sqrt{5^{2}+2^{2}} \times \cos \theta$

$\text { since }\|a\|=\sqrt{x^{2}+y^{2}} \text { for } a=x \vec{\imath}+y \vec{\jmath}$

$\begin{array}{l}{-10-10=\sqrt{29} \times \sqrt{29} \times \cos \theta} \\\\ {-20=29 \times \cos \theta} \\\\ {\cos \theta=\frac{-20}{29}} \\\\ {\theta=\cos ^{-1} \frac{-20}{29}} \\\\ {\theta=133.60}\end{array}$

Hence, the angle between given two vectors is 134 degrees approximately.

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Step-by-step explanation:

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