All categories

3 years ago

What is the slope of the line that passes through the points (5,3) and (-4,1)

Mathematics

1 answer:

wariber [46]3 years ago

5 0

Answer:

$\frac{2}{9}$

Step-by-step explanation:

Formula to find the slope: $\frac{y_{2}-y_1}{x_2-x_1}$

Notice it doesn't matter for which point being y2/x2 or y1/x1.

Just to make our life easier, I choose (5,3) as our second point and (-4,1) as our first point.

$\frac{3-1}{5-(-4)}$

= $\frac{2}{9}$

You might be interested in

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$

and this converges to $g(0,0)=0$ , since differentiability of $g$ means

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0$

So, $f$ is differentiable at (0, 0).

3 0

3 years ago

Considering the activity series given below for metals and nonmetals, which reaction will occur? al > mn > zn > cr >

victus00 [196]

The activity series is an arrangement of species in order of reactivity.

<h3>What is the activity series?</h3>

The activity series is an arrangement of species in order of reactivity. The position of a specie in the series determines the possibility of certain reactions.

The following are the fates of these reactions;

2NaBr + I2--------->2NaI + Br2 - Not possible
2Fe + Al2O3 ----------> 2Al + Fe2O3 - Not possible
2AgNO3 + Ni ----------> Ni(NO3)2 + 2Ag - possible
Pb + Zn(C2H3O2)2 ----------> Zn + Pb(C2H3O2)2 - Not possible

Learn more about activity series:brainly.com/question/18080147

#SPJ1

8 0

1 year ago

Please help me solve this.<br> 15x+8=-2+14x

Airida [17]

Answer:

x=-10

Step-by-step explanation:

15x+8=-2+14x

15x-14x=-2-8

x=-10

5 0

3 years ago

Help please logarithms

DochEvi [55]

Answer:

Simplified: 40

Step-by-step explanation:

To simplify this, we need to first separate the terms in this expression. There is a property of exponents that says: $x^{a+b} =x^ax^b$ We can do this same thing to simplify this expression. Next, we can simplify each of these terms. $2^2$ simply becomes 4. In the other term, the $2^{log_2}$ simply cancel each other and leave the 10. This means that we're left with 4*10, which is 40

$2^{2+log_210} \\\\2^2*2^{log_210} \\\\4*10\\\\40$

6 0

2 years ago

in the previous question, what is the distance from a side of the triangle to the incenter? can someone also explain how to get

denis23 [38]

In centre is the centre of a circle that is inscribed in a triangle, called the incircle, i.e. circle is tangent to all three sides AND located inside the triangle. An incircle is roughly sketched in the accompanying diagram.

Since all radii of an incircle are equal, we have
4x-1=6x-5
Solving for x,
6x-4x=5-1
x=2
consequently, distance from incentre to each side of the triangle is
4(2)-1=7
check 6x-5=6(2)-5=7... checks.

Answers:
The value of x = 2
The distance from incentre to side of triangle is 7

5 0

2 years ago

What is the slope of the line that passes through the points (5,3) and (-4,1)​

What is the slope of the line that passes through the points (5,3) and (-4,1)