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wel
3 years ago
6

Is the triangle with the side lengths 5,12,13 right, obtuse, or acute? Why?

Mathematics
1 answer:
Art [367]3 years ago
5 0
Acute because it is not large enough to be an obtuse or not perfect to be a rught
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Can i plz get some help with this!<br>its about finding the conjecture<br>​
Studentka2010 [4]

Answer:

3-they are cubes of consecutive natural numbers

1 cube is 1

2 cube is 8

3 cube is 27

and so on...

let me know if you're asking fourth one or third one

6 0
3 years ago
Please help me !!!!!!!
erastova [34]

Answer:

C. Both (1,11) and (-1,-5)

Step-by-step explanation:

Ok so,

let's try to plug in the first ordered pair into the equation. It should look like this:

11 = 8+3

11=11      CORRECT

now let's try to plug in the second ordered pair. it should look like this:

-5 = -8+3

-5 = -5      CORRECT

hope this helps. have a good day!!

6 0
3 years ago
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side
lorasvet [3.4K]

Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

       =cot x (sec²x)²

       =cot x (1+tan²x)²     [ ∵ sec²x=1+tan²x]

       =  cot x(1+ 2 tan²x +tan⁴x)

       =cot x+ 2 cot x tan²x+cot x tan⁴x

        =cot x +2 tan x + tan³x        [ ∵cot x tan x =\frac{ \textrm{tan x }}{\textrm{tan x}} =1]

       =R.H.S

2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

 L.H.S =(sin x)(tan x cos x - cot x cos x)

          = sin x tan x cos x - sin x cot x cos x

           =\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}

           = sin²x -cos²x

           =1-cos²x-cos²x

           =1-2 cos²x

           =R.H.S

         

3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

         =1+\frac{{sin^2x}}{cos^2x}                       [\textrm{sec x}=\frac{1}{\textrm{cos x}}]

         =1+tan²x                        [\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]

         =sec²x

        =R.H.S

4.

\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}

L.H.S=\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}

       =\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}

      =\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}

     =\frac{\textrm{2sin x}}{sin^2 x}

      = 2 csc x

    = R.H.S

5.

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

        = sec²x-tan²x

        =\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}

        =\frac{1- sin^2x}{cos^2x}

        =\frac{cos^2x}{cos^2x}

        =1

     

       

8 0
3 years ago
In a right triangle ,one acute angle measure 39, what is measure of the other acute angle?
LUCKY_DIMON [66]
C, 51 because acute angles are less than 90° but together the angles create a right triangle which is 90°.
3 0
3 years ago
Read 2 more answers
What is the product of the polynomials below?
Anna35 [415]
The answer is letter C
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