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Natali [406]
3 years ago
11

The slope that goes through the points (3,-4) and (-2,6)​

Mathematics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

slope =  -2

Step-by-step explanation:

Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

         = \frac{6-[-4]}{-2-3}\\\\= \frac{6+4}{-2-3}\\\\= \frac{10}{-5}\\\\= -2

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Rex paid $250 in taxes to the school district that he lives in this year. This year's taxes were a 12% increase from last year.
Verizon [17]

Answer:

$223.21

Step-by-step explanation:

Last year the amount of taxes was x, an unknown amount. That was 100% of last year's tax cost.

This year, taxes went up by 12%, so this year the taxes are 112% of what they were last year. This year the taxes were 112% of x, and they were $250.

112% of x = 250

1.12x = 250

x = 250/1.12

x = 223.21

Answer: $223.21

8 0
3 years ago
Kim has 5 yards of denim. She cuts each yard of denim into thirds. How many 1/3-yard pieces of denim does she have?
NNADVOKAT [17]

Answer:

15

Step-by-step explanation:

1/3(15)

15/3=5

3 0
3 years ago
Read 2 more answers
Graphing rational functions. I need help asap!!!!
Anastaziya [24]

Answer:

1.  

Vert. asymptote: x = {-3, 2}

Horiz. asymptote: y = 0

x-int: None

Question 3.

a. There is no hole

b. Vert. asymptote: x = {-2, 2}

c. f(x) = 0: x = {0, -1/2}

d. The graph has no hole at (-2, 4)

Question 4.

a. Vert. asymptote: x = {-2, 2}

b. f(x) = 0: x = {0, -1/2}

c. Horizontal asymptote: y = 2

d. The graph has no hole

I'm a bit confused.  Some of the things stated in the question aren't true like how there are holes in places where there aren't.

6 0
3 years ago
Find the exact value of the trigonometric expression.
Dmitrij [34]
\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}

\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}

\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}


which simplifies thus far to 

\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}

\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}
7 0
3 years ago
Which graph represents the solution set of the inequality x + 2 greater-than-or-equal-to 6
Tanzania [10]

Answer:

  x ≥ 4; solid circle at 4 and shaded to the right

Step-by-step explanation:

Given:

  x +2 ≥ 6

Subtract 2

  x +2 -2 ≥ 6 -2

  x ≥ 4

This is graphed as a solid circle at x=4 and shaded to the right (where x > 4).

5 0
3 years ago
Read 2 more answers
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