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Fittoniya [83]
3 years ago
6

Resuelve las siguientes ecuaciones tales que 0° ≤ x ≤ 360°

Mathematics
1 answer:
Vladimir79 [104]3 years ago
4 0

Answer:

4cos=2X

X=3-4COS

X=-1

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Which graph represents the function?
BlackZzzverrR [31]

Answer:

The last one.

Step-by-step explanation:

The solutions with the horizontal lines are wrong because then h(x) would be -1 and 1 respectively.

The subtlety is in the open or closed dot. x ≥ -1 means that -1 is included, which is drawn as a closed dot. x < -1 means that -1 is excluded, which is drawn as an open dot.

Hence the last graph is the correct one.

4 0
3 years ago
Read 2 more answers
Select the correct answer from the drop-down menu.
lidiya [134]

Answer:

6y² - 20y + 6

Step-by-step explanation:

Use FOIL method

(3y - 1)(y-3) = 3y*y + 3y*(-3) + (-1)*y + (-1)*(-3)

                 = 3y²- 9y - y + 3    {add the like terms}

                = 3y² - 10y + 3

2(3y - 1)(y - 3) = 2 ( 3y² - 10y + 3)

                      = 2*3y² - 2*10y + 2*3

                      = 6y² - 20y + 6

6 0
3 years ago
Determine the slope for the line below
kow [346]

Answer:

C : 20

good luck, hope this helped!

7 0
3 years ago
Evaluate the expression. -38 + 11<br> a. -27<br> b. -49<br> c. 49<br> d. 27
kiruha [24]
The answer is a.) -27

subtract 11 from 38
27 keeps the negative sign because -38 is larger than 11
8 0
3 years ago
PLEASE HELP ASAP! NO SCAMS ALLOWED!
Kitty [74]

Answer:

Step-by-step explanation:

I used calculus for this, as I'm not sure there's any other way to do it and to do it as easily. This is the volume of a solid found by using the disk method of rotation:

V=\pi\int\limits^a_b {[R(x)]^2-[r(x)]^2} \, dx

where R(x) is the outer shell of the solid and r(x) is the space inbetween the solid and the axis of rotation. There is no space between the solid and the axis of rotation, so r(x) = 0. R(x) is the height of the solid which is 3. Therefore, f(x) = 3 and that's the function we put into the formula with the lower bound of 3 and the upper bound of 5:

V=\pi\int\limits^5_3 {3^2}-0^2 \, dx  and

V-\pi\int\limits^5_3 {9} \, dx  and integrating:

V=\pi(9x\left \} {{5} \atop 3}} \right. and using the First Fundamental Theorem of Calculus:

V = π(9(5) - 9(3)) and

V = π(45 - 27) so

V = 18π units cubed or in decimal format,

V = 56.549 units cubed

6 0
3 years ago
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