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3 years ago

What is the slope of the line that passes through the points (4, -6) and (10, 12)?

2 answers:

4 0

We know the equation has to be in the form y=mx+c

12--6/10-4
=18/6
=3 (this 3 is our m value)
now we know the gradient we can substitute in one of the points

y=3x+c
12=(3×10)+c
12=30+c
-18=c

therefore the equation of line is y=3x-18

Vitek1552 [10]3 years ago

3 0

The equation for finding the slope of a line is $\frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }}$ .

Inserting the points, we get $\frac{ -6 - 12 }{ 4 - 10 }}$ which is 6 / 6 or 1.

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Jake ate 55% of a cake. What fraction of the cake did he eat?

7nadin3 [17]

Answer:

11/20

Step-by-step explanation:

To be able to find the fraction of the cake that Jake ate if you know that he ate 55% of it, you have to turn the 55% into a fraction. To do this, first you have to divide the percent by 100:

55/100

Then, you have to simplify the fraction. In this case, you can do it by dividing by 5:

11/20

According to this, the fraction of the cake that Jake ate is 11/20.

8 0

3 years ago

Triangle ABC has points at A(2, 0), B(−1, 3), and C(4, 5). After undergoing a transformation, its new points are A'(6, 0), B'(3,

djverab [1.8K]

Answer:C

Step-by-step explanation:

6 0

3 years ago

Elyria Warehousing desired to locate a central warehouse to serve five Ohio markets. Placed on a grid system, its five markets h

Iteru [2.4K]

Answer:

The correct option is 2.

Step-by-step explanation:

According to the the center-of-gravity technique, the coordinates of the center-of-gravity location are

$(\frac{\sum x_iL_i}{\sum L_i},\frac{\sum y_iL_i}{\sum L_i})$

Where ( $(x_i,y_i)$ represent the coordinates and $L_i$ is demand.

We have to find the Y-coordinate of the center-of-gravity location.

The sum of product of demand and corresponding y coordinates is

$\sum y_iL_i=65\times 2200+55\times 900+95\times 1300+200\times 1750+175\times 3100=1208500$

The sum of demanded units is

$\sum L_i=2200+900+1300+1750+3100=9250$

The Y-coordinate of the center-of-gravity location is

$y_0=\frac{\sum y_iL_i}{\sum L_i}$

$y_0=\frac{1208500}{9250}$

$y_0=130.6486$

$y_0\approx 131$

The Y-coordinate of the center-of-gravity location is 131. Therefore the correct option is 2.

7 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago

Solve the algebraic expression below (2/5) (10) + 8

Andrews [41]

Answer: 12.

Step-by-step explanation:

4 0

3 years ago