What is the slope of the line through (3, 2) and (-3, 4)?

Mathematics

1 answer:

OlgaM077 [116]2 years ago

3 0

Answer:

- 1/3

Step-by-step explanation:

You might be interested in

HELP ME ASAP PLEASE!!

lilavasa [31]

Answer:

a=45 b=65

Step-by-step explanation:

You know that all the angle measures of a triangle add to 180. So the equation would be angle a + angle b + 70 = 180. Subsitute and simplify and you get 16x -2 + 70 = 180. Subtract 70 from 180 to get 16x -2=110 add 2 to get 16x =110. Divide each side by 16 to gte x=7. Subsitute x = 7 into the equations for angle a and angle b to get the answers above

4 0

2 years ago

The first term of an arithmetic sequence is -5, and the tenth term is 13. Find the common difference.

zzz [600]

Hello,

$u_{1} =-5\\
 u_{2} =-5+d\\
 u_{3} =-5+2d\\
 u_{4} =-5+3d\\
...\\
 u_{10} =-5+9d=13\\
==\ \textgreater \ 9d=18 \\
==\ \textgreater \ d=2

$

6 0

3 years ago

I don’t bought a new jacket at 75% off it’s original price was $80 and the sales tax is 15% what is the total cost

Vika [28.1K]

Answer:

$69

Step-by-step explanation:

80 x 75% (0.75) = 60

60 x 15% (0.15) = 9

60 + 9 = $69

5 0

2 years ago

Let A and B be subsets of R. (a) If x ∈ (A ∩ B)c, explain why x ∈ Ac ∪ Bc. This shows that (A ∩ B)c ⊆ Ac ∪ Bc. 12 Chapter 1. The

cupoosta [38]

Answer:

answer is -3 just subtract 4 from each side

Step-by-step explanation:

7 0

2 years ago

If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?

Keith_Richards [23]

$\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}$

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

$\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}$

so, the domain is all real numbers EXCEPT that one.

4 0

3 years ago