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prohojiy [21]
4 years ago
12

How are points graphed on a coordinate plane?

Mathematics
2 answers:
bogdanovich [222]4 years ago
8 0
Points are graphed on a coordinate plane by placing a dot on a location on a graph.
amid [387]4 years ago
3 0
Points are graphed like this: if you had the ordered pair (2,3), first you would go to the Y axis (vertical), then move to the 3 on the X Axis (horizontal) from where the 2 is.
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If you made 20$ a week and your goal was 200$ how many days would it take to reach your goal?
wolverine [178]
It would take you 70 days.  
4 0
3 years ago
Read 2 more answers
P is a prime number and q is a positive integers such that p + q = 1696 IF P and Q are co primes and their Lcm is 21879 Then fin
Stolb23 [73]

Answer:

P = 1 3  

Q = 1 6 8 3

Step-by-step explanation:

through factorization of 21879

4 0
3 years ago
Explain the difference between using the tangent ratio to solve for a missing angle in a right triangle versus using the cotange
Lena [83]

Answer:

While determining the angle \theta with the tangent ratio or cotangent ratio uses the same sides lengths but the ratios are inverse of each other.

Step-by-step explanation:

See the diagram attached.

Let us assume a right triangle Δ ABC with ∠ B = 90°. Now, assume that the angle ∠ CAB = \theta and the sides AB, BC, CA are 3, 4, 5 units respectively.

The tangent ratio of an angle \theta is given by  

\tan \theta = \frac{BC}{AB} = \frac{4}{3}  

Again, the cotangent ratio of angle \theta is given by  

\cot \theta = \frac{AB}{BC} = \frac{3}{4}  

Therefore, in both the cases of tangent ratio and cotangent ratio are inverse of each other and while determining the angle \theta with the tangent ratio or cotangent ratio uses the same sides lengths but the ratios are inverse of each other. (Answer)

4 0
4 years ago
PLEASE HELP THE ANSWER HAS TO BE A INTEGER OR A DECIMAL.
svlad2 [7]

Answer:

It's the same as the last one. The answer is -2.3

Step-by-step explanation:

3 0
3 years ago
Select the correct answer from each drop-down menu. The equation (y-2)^2/3^2 - (x-2)^2/4^2=1 represents a hyperbola whose foci a
aev [14]

Answer:

The foci are (2 , 7) and (2 , -3)

Step-by-step explanation:

* lets revise the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- The coordinates of the vertices are ( h ± a , k )  

- The coordinates of the co-vertices are ( h , k ± b )

- The coordinates of the foci are (h , k ± c), where c² = a² + b²

* Now lets solve the problem

∵ The equation of the hyperbola of vertex (h , k) is

    (y - k)²/a² - (x - h)²/b² = 1

∵ The equation is (y - 2)²/3² - (x - 2)²/4² = 1

∴ k = 2 , h = 2 , a = 3 , b = 4

∵ The foci of it are (h , k + c) and (h , k - c)

- Lets find c from the equation c² = a² + b²

∵ a = 3

∴ a² = 3² = 9

∵ b = 4

∴ b² = 4² = 16

∴ c² = 9 + 16 = 25

∴ c = √25 =  5

- Lets find the foci

∵ The foci are (h , k + c) and (h , k - c)

∵ h = 2 , k = 2 , c = 5

∴ The foci are (2 , 2 + 5) and (2 , 2 - 5)

∴ The foci are (2 , 7) and (2 , -3)

5 0
4 years ago
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