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irga5000 [103]
3 years ago
14

Find the measure of m<HMG

Mathematics
1 answer:
Nookie1986 [14]3 years ago
8 0
Do u have an image ? is there a picture?

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Translate the following problem into an equation: (1) together Jon and Amy spent $29 dollars on a birthday gift. (2) Jon spent $
kolbaska11 [484]
Let the money spent by John = y
and Amy=x

ATQ

x+y = $29

It is given that jon spend 5 more than Amy. SO he spend

x +5

x+y = $29
x+x+5=$29

2x = 29-5

2x= 24

x = 24/2 =12

So Amy spend $ 12 and Jon spend x+5 = 12+5 = $17
4 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
A flagpole broke in a storm. It was originally 81 8181 feet tall. 28 2828 feet are still sticking straight out of the ground, wh
solong [7]

Answer:

The end of the flagpole is 50.79 ft away from the base of the pole.

Step-by-step explanation:

The problem is represented by the diagram below.

The broken flagpole forms the shape of a right angled triangle. We need to find one of the sides of the triangle, the adjacent (x).

The hypotenuse is the broken part of the flagpole (53 ft), while the opposite is the part of the flagpole that is still stuck to the ground (28 ft).

Using Pythagoras theorem, we have that:

hyp^2 = adj^2 + opp^2

=> 53^2 = x^2 + 28^2

3364 = x^2 + 784\\\\=> x^2 = 3364 - 784\\\\x^2 = 2580\\\\x = \sqrt{2580}\\ \\x = 50.79 ft

The end of the flagpole is 50.79 ft away from the base of the pole.

7 0
3 years ago
Can some one help, step bye step please
il63 [147K]

Answer:

$14.23

Step-by-step explanation:

Multiply the original value by 1.035 to get your answer.

4 0
3 years ago
Read 2 more answers
Consider the infinite geometric series 2(p-5) + 2(p-5)² + 2(p-5)³​
nasty-shy [4]
10(p+3)-4(p+10)-6(p+15)
7 0
3 years ago
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