All categories

3 years ago

Can anyone help me? My teacher just recently taught me this today, be he never explained much. The question is in the photo

Mathematics

1 answer:

IceJOKER [234]3 years ago

4 0

You've got five different problems in this photo ... four on top and the word problem on the bottom ... and they're all exactly the same thing: Taking two points and finding the slope of the line that goes through them.

In every case, the procedure is the same.
If the two points are (x₁ , y₁) and (x₂ , y₂) , then
the slope of the line that goes through them is

      Slope = (y₂ - y₁) / (x₂ - x₁) .

This is important, and you should memorize it.

#1). (8, 10) and (-7, 14)

   Slope = (14 - 10) / (-7 - 8) = 4 / -15

#2). (-3, 1) and (-17, 2)

   Slope = (2 - 1) / (-17 - -3) = (2 - 1) / (-17 + 3) = 1 / -14

#3). (-20, -4) and (-12, -10)

   Slope = [ -10 - (-4) ] / [ -12 - (-20) ]

=========================================

The word problem:

This question only gives you one point on the graph,
and then it wants to know what's the slope ?
What are you going to do for another point ?

A "proportional relationship" always passes through the origin,
so another point on the line is (0, 0) .

Now you have two points on THAT line too, and you can easily
find its slope.

You might be interested in

Please anwser

Black_prince [1.1K]

The answers are
M1 = 112°
M2 = 26°

6 0

2 years ago

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v

olga_2 [115]

We are given

$f=z^3 -x^2y$

Firstly, we can find gradient

so, we will find partial derivatives

$f_x=0 -2xy$

$f_x=-2xy$

$f_y=0 -x^2$

$f_y=-x^2$

$f_z=3z^2$

now, we can plug point (-5,5,2)

$f_x=-2*-5*5=50$

$f_y=-(-5)^2=-25$

$f_z=3(2)^2=12$

so, gradient will be

$gradf=(50,-25,12)$

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

$|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}$

$|v|=\sqrt{29}$

now, we can find unit vector

$v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

now, we can find dot product to find direction of the vector

$dir=(gradf) \cdot (v')$

now, we can plug values

$dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

$dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })$

$dir=-\frac{248\sqrt{29}}{29}$ .............Answer

7 0

3 years ago

Read 2 more answers

Find the values of a through e that make these two relations inverses of each other.

yuradex [85]

Answer:

hmm I'm pretty sure a is 3.2

4 0

11 months ago

Helppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

Anna [14]

Answer:

are you ok?

Step-by-step explanation:

why did you need help lol

5 0

3 years ago

Read 2 more answers

Rename the number to 120,000 ten thousand

Ray Of Light [21]

120,000 renamed to ten thousand (10,000) is twelve ten thousand. Ten thousand in number form is 10,000. When you divide 120,000 (one hundred and twenty thousand) by 10,000 (ten thousand) you get 12. 120,000 (one hundred and twenty thousand) /10,000 (ten thousand) = 12. So there are twelve ten thousands in 120,000.

4 0

2 years ago