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valkas [14]
3 years ago
10

Find mCE A. 40 DEGREES B. 50 DEGREES C. 75 DEGREES D. 80 DEGREES

Mathematics
1 answer:
nordsb [41]3 years ago
3 0
B. 50 degrees hope this helps
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Glass melts at a temperature that is higher than 2,600 degrees and less than 2,900 degrees.​
Agata [3.3K]

Answer:

2,700-2,800

Step-by-step explanation:

3 0
3 years ago
Quadrilateral ABCD with vertices A(-4, 1), B(-2,3), C(0,-2), and D(-5,-2) Give the coordinates of the image if k=3.
borishaifa [10]

If you're using a coordinate plane, I think it makes a trapezoid

3 0
3 years ago
Help meee I’ll give 10 pts and brainliest!!!
crimeas [40]

Step-by-step explanation:

i) \overline{AB} = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}

\:\:\:\:\:\:\:=\sqrt{(2)^2 + (12)^2} = 12.3

ii) m = \dfrac{y_A - y_B}{x_A - x_B} = \dfrac{-12}{2} = -6

iii) (\overline{x},\:\overline{y}) = \left(\dfrac{x_A + x_B}{2},\:\dfrac{y_A + y_B}{2}\right)

\:\:\:\:\:\:\:=(3,\:-2)

7 0
3 years ago
Write and equation of the translated or rotated graph in general form (picture below)
WINSTONCH [101]

Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

 xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

  the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

- The red line is x'

- The blue line is y'

6 0
3 years ago
Read 2 more answers
4p+6-3 combine like terms?
amm1812

Answer:

4p+3

Step-by-step explanation:

Because 4 has the variable p next to it, 4 can not be the same as 6 or 3 because they are whole numbers. So the only thing you can do here is 6-3, which is 3.

4 0
2 years ago
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