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Mashcka [7]
3 years ago
5

2cot^2(x)-7csc(x)+8=0

Mathematics
1 answer:
Crank3 years ago
6 0
Hello here is a solution : 

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50 points!!!Adam recorded the number of flower blooms on each of his ten rose bushes. 9, 10, 4, 12, 15, 17, 18, 10, 12, 13If he
lyudmila [28]

Answer:

12

Step-by-step explanation:

9, 10, 4, 12, 15, 17, 18, 10, 12, 13

Putting the numbers from smallest to largest

4,9, 10, 10, 12  12, 13, 15, 17, 18,

The median is the middle number

There are 10 numbers

4,9, 10, 10, 12          12, 13, 15, 17, 18,

The middle number is between the 12's, so it is 12

6 0
3 years ago
Read 2 more answers
Find the central angle of a sector of a circle if the area of the sector and the area of the remaining part of the circle are in
andre [41]

Answer:

288

Step-by-step explanation:

360/5 = 72

72x4 = 288

8 0
3 years ago
Secants L J and L M intersect and form an angle at point L. Solve for x.
german

Answer:

x = 14  

Step-by-step explanation:

Assume your diagram is like the one below.

The intersecting secant angles theorem states, "When two secants intersect outside a circle, the measure of the angle formed is one-half the difference between the far and the near arcs."

For your diagram, that means

\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}

Check:

\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}

It checks.

6 0
3 years ago
Plz help me on 8 question
Salsk061 [2.6K]
No they are not at $3 for 6 bagles for $9 you only get 18 BAGLES
8 0
3 years ago
Change 20% into a fraction
OlgaM077 [116]
I think it’s just 20/100
5 0
3 years ago
Read 2 more answers
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