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ivolga24 [154]
2 years ago
10

The point slope equation of a line is

Mathematics
2 answers:
vivado [14]2 years ago
5 0

Answer:

y-y1=m (x-x1)

Step-by-step explanation:

I hope this is what you're looking for!

EastWind [94]2 years ago
3 0

Step-by-step explanation:

Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept).

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How do I solve for x??
Cloud [144]

The two angles given are vertical angles which mean they are the same.

x +40 = 60

Subtract 40 from both sides:

x = 20

4 0
3 years ago
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In an arithmetic sequence, term 10 is 33 and term 22 is -3. What are the first 4 terms
Zarrin [17]

Answer:60,57,54,51

Step-by-step explanation:

7 0
3 years ago
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Verify the identity. 4 csc 2x = 2 csc2x tan x
vlada-n [284]

Step-by-step explanation:

4csc(2x) = 2csc^2(x) tan(x)

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

4 \frac{1}{sin(2x)}

Also we use identity

sin(2x) = 2 sin(x) cos(x)

4 \frac{1}{2sin(x)cos(x)}

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}

2\frac{sin(x)}{sin^2(x)cos(x)}

2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}

We know that sinx/ cosx = tan(x)

Also  1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified



6 0
3 years ago
Find cot θ if csc θ = negative square root of thirty seven divided by six and tan θ > 0.
Ket [755]
You can use the trigonometric identity  1+\cot^2{x}=\csc^2{x}.

1+\cot^2{\theta}=(\frac{-\sqrt{37}}{6} )^2 \\ 1+\cot^2{\theta}=\frac{37}{36} \\ \cot^2{\theta}=\frac{1}{36} \\
\cot{\theta}=\frac{1}{6}

The requirement that \tan{\theta}\ \textgreater \ 0 eliminates -1/6 from being another solution.
6 0
2 years ago
Find the percent change from 70 to 42.
Evgen [1.6K]

Answer:

It decreased by 40%

Step-by-step explanation:

Percent change = 42 - 70

|70|

x 100 = -28

70

x 100 = -40

Mark me brainliest if I helped:D

3 0
2 years ago
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