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Dmitry [639]
3 years ago
8

Prove that . 1+cot2 theta = cosec2theta​

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
6 0

Answer:

see explanation

Step-by-step explanation:

Using the Pythagorean identity

sin²θ + cos²θ = 1 ( divide terms by sin²θ )

\frac{sin^20}{sin^20} + \frac{cos^20}{sin^20} = \frac{1}{sin^20} , that is

1 + cot²θ = cosec²θ ← as required

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One day, the temperature started at 12 degrees at 6:00 a.m., then climbed 3 degrees by noon, and then dropped 5 degrees by midni
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4 years ago
8w-6(w-4)=2(w+9) <br> please help asap
uranmaximum [27]

Answer:

w = -7.04 and 1.705

Step-by-step explanation:

First lets distribute.

We have 8w-6w(w-4), so let's distribute the -6w to w and -4

8w-6w^2+24w = 2(w+9)

Now for the other side we distribute 2 to w and 9

8w-6w^2+24w = 2w +18

Now we combine like terms

6w^2+32w = 2w +18

Carry both 2w and 18 to the other side

6w^2 +30w -18 = 0

Now we can factor the equation, i used the quadratic method which is x= -b+- the square root of b^2-4ac /2a (in this case x would be w)

Plugging that in:

x= -32 +- square root of 32^2 -4(6 * -18) / 2(6)

-32 +- square root of 1024 +1728 / 12

-32 +- square root of 2752 /12

-32 +- 52.46 /12

x = -32 - 52.46/12   and  -32 + 52.46/12

x=  -84.46/12      and      20.46/12

x= -7.04  and  1.705

so w = -7.04 and 1.705

Sorry this was so complicated, but I hope this helps

6 0
2 years ago
Use the discriminant to determine the number and type of solutions for the following equation. 12x2 + 10x + 5 = 0
Paul [167]
Discriminant D is given by:
D=b²-4ac
Implication of discriminant is as follows:

D<0 two zeros that are complex conjugate
D=0 one real zero of multiplicity 2
D>0 two distinct real zers
D= (+ve perfet square)  two distinct rational zeros
From:
12x^2+10x+5=0
plugging in the equation we get:
10²-4×12×5
=100-240
=-140
thus
D<0

Answer is: 
<span>A two irrational solutions </span>
4 0
4 years ago
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artcher [175]

either a rectangle or circle will be made. If you cut it horizontially you will get a circle. Vertically and you will have a rectangle

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