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OverLord2011 [107]
3 years ago
10

Find the slope of a line that passes through the points (5,4) and (-2,-1)

Mathematics
1 answer:
Anna11 [10]3 years ago
7 0
The slope is 5/7
4-(-1)=5 which is the Y value
5-(-2)=7 which is the X value
Slope =Y/X
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The Answer to This Question
xenn [34]

Recall the Trigonometric Ratio below:

\tt{ \large{sin \theta =  \frac{opposite}{hypotenuse} } }\\   \tt{\large{cos \theta =  \frac{adjacent}{hypotenuse} }} \\    \tt{ \large{tan \theta =  \frac{opposite}{adjacent} }}

For csc, sec and cot - they are reciprocal of sin,cos and tan.

\tt{ \large{csc  \theta =  \frac{1}{sin \theta} } }\\  \tt{ \large{sec \theta =  \frac{1}{cos \theta} }} \\  \tt{ \large{cot \theta =  \frac{1}{tan \theta} }}

What we know now is our hypotenuse, adjacent and opposite length.

  • hypotenuse = 15
  • opposite = 12
  • adjacent = 9

Therefore,

\large{sin \theta =  \frac{12}{15}  \longrightarrow  \frac{4}{5} } \\ \large{cos \theta =  \frac{9}{15}   \longrightarrow  \frac{3}{5} } \\  \large{ tan \theta =  \frac{12}{9}  \longrightarrow  \frac{4}{3} }As for the reciprocal of three trigonometric ratio. We just swap the numerator and denominator.

\large{csc \theta =  \frac{15}{12}  \longrightarrow  \frac{5}{4} } \\  \large{sec \theta =  \frac{15}{9}  \longrightarrow  \frac{5}{3} }   \\  \large{cot \theta =   \frac{9}{12}  \longrightarrow \frac{3}{4} }

Answer

  • sin = 12/15 —> 4/5
  • cos = 9/15 —> 3/5
  • tan = 12/9 —> 4/3
  • csc = 15/12 —> 5/4
  • sec = 15/9 —> 5/3
  • cot = 9/12 —> 3/4

The first is non-simplifed form while the second that has the arrow pointing is the simplest form.

Let me know if you have any doubts.

5 0
3 years ago
Read 2 more answers
Solve for x! Someone helppp
olganol [36]

Answer:

17.2

Step-by-step explanation:

8.7x-1.9x=116.96

6.8x=116.96

x=116.96/6.8

x= 17.2

6 0
2 years ago
The sum measures of angle X and angle Y is 90. If the angle X is 30 less than twice the measure of angle Y, what is the measure
krek1111 [17]

Answer:

C. 50

Step-by-step explanation:

x+y=90

x=2y-30

90-y=2y-30

3y=120

y=40

90-40= 50


6 0
4 years ago
If the quadratic function f(x)= ax^2+bx+c produces a parabola with the vertex on the x-axis which of the following statements is
loris [4]
It would have to be B, because with a vertex on the x-axis, the only point it would touch the x-axis is the vertex, thus proving that this quadratic has only one zero.
6 0
3 years ago
F (x)= 6x^2 + 6x + 12 (Solve by factoring.)
zzz [600]
You can't solve for the zeros by factoring because the function is not factorable.

I'll show you what happens:
f(x) = 6x^2 + 6x + 12

To solve for the zeros (when f(x)=0), set f(x) equal to 0.
0 = 6x^2 + 6x + 12

First you can factor out a 6.
0 = 6(x^2 + x + 2)

Divide by 6.
0 = x^2 + x + 2

Now you need two integers that add up to the coefficient of x (1) and multiply to the product of the last term and the coefficient of x^2. This number is 2.
No two integers add up to 1 and multiply to 2.
3 0
3 years ago
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