All categories

3 years ago

The slope of the line passes through the points (2,3) and (5,9) is

Mathematics

2 answers:

timurjin [86]3 years ago

4 0

Answer:

Step-by-step explanation:

m = y2-y1/x2-x1

m = 9-3/5-2

m = 6/3

m = 2

wolverine [178]3 years ago

3 0

Slope = 1/2

hope this helps

You might be interested in

Find the slope of the line with these two pair of points (10,-5) and (8,-7)

matrenka [14]

Answer:

Step-by-step explanation:

8 0

3 years ago

Lim x->pi/3 ((2cosx-1)/(tan^2x-3))

beks73 [17]

You can check that the limit comes in an undefined form:

$\displaystyle \lim_{x\to\frac{\pi}{3}} \frac{2\cos(x)-1}{\tan^2(x)-3}=\dfrac{0}{0}$

In these cases, we can use de l'Hospital rule, and evaluate the limit of the ratio of the derivatives. We have:

$\dfrac{\text{d}}{\text{d}x}2\cos(x)-1 = -2\sin(x)$

and

$\dfrac{\text{d}}{\text{d}x}\tan^2(x)-3 = 2\dfrac{\tan(x)}{\cos^2(x)}=\dfrac{2\sin(x)}{\cos^3(x)}$

So, we have

$\displaystyle \lim_{x\to\frac{\pi}{3}} \frac{2\cos(x)-1}{\tan^2(x)-3}=\lim_{x\to\frac{\pi}{3}} \dfrac{-2\sin(x)}{\frac{2\sin(x)}{\cos^3(x)}}=\lim_{x\to\frac{\pi}{3}}-\cos^3(x)=-\cos^3\left(\dfrac{\pi}{3}\right)$

6 0

3 years ago

Find the circumference and the area of a circle with radius 9<br> .

Eduardwww [97]

C=2(pi)r use this equation with 9 in place of r

6 0

3 years ago

The Pythagorean theorem is a^2+b^2=c^2<br> Solve for b.

rodikova [14]

Answer:

see explanation

Step-by-step explanation:

given

a² + b² = c² ( isolate b² by subtracting a² from both sides )

b² = c² - a² ( take the square root of both sides )

$\sqrt{b^{2} }$ = ± $\sqrt{c^2-a^2}$

b = ± $\sqrt{c^2-a^2}$

3 0

3 years ago

Read 2 more answers

Given: Line A R bisects ∠BAC; AB = AC Triangle A B C is shown. Point R is at the middle of the triangle. Lines are drawn from po

Dahasolnce [82]

Answer:

It is SAS

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers