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wolverine [178]
3 years ago
8

Identify the percent of change an increase or decrease then find percent of change one hundredfifty pounds to one hundred thirty

five pounds
Mathematics
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

−10%change

 10%decrease

Step-by-step explanation:

V2−V1)|V1|×100

 =(135−150)|150|×100

=−15150×100

=−0.1×100

=−10%change

=10%decrease

Note: Percent Change is NOT the same as  Percent Difference between 150 and 135.

Hope this helps!

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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
Building a is 160160 feet shorter than building
Serjik [45]
I assume the heights are 160 ft and 1480 ft.

The two heights are unknown, so we will use variable h to help solve the problem.
The shorter building, building A, has height h.
Since building A is shorter by 160 ft, then building B is taller by 160 ft, so the height of building B is h + 160.

Now we add our two heights to find the total height.

h + h + 160 is the total height.
We can write it as 2h + 160

We are told the total height is 1480 ft, so we let 2h + 160 equal 1480, and we have an equation.

2h + 160 = 1480

Subtract 160 from both sides

2h = 1320

Divide both sides by 2

h = 660

h + 160 = 820

Building A measures 660 ft.
building B measures 820 ft.
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Prestige Car Rentals charges $44 per day plus 6 cents per mile to rent a mid-sized vehicle. Gateway Auto charges $35 per day plu
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Make two equations from the information. Then graph them and see where they intersect. The answer is 62 miles. Plug 62 back into both equations and they should both be the same number.

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padilas [110]

Answer:

d=7.07

Step-by-step explanation:

Distance Formula: d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>:

d=\sqrt{(3-(-2))^2+(-4-1)^2}

d=\sqrt{(3+2)^2+(-5)^2}

d=\sqrt{(5)^2+25}

d=\sqrt{25+25}

d=\sqrt{50}

d=5\sqrt{2}

d=7.07

5 0
3 years ago
The area of a regular octagon is 35 cm squared what is the area of a regular octagon with sides five times as long
KATRIN_1 [288]
The answer would be 875cm squared
7 0
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