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Hoochie [10]
2 years ago
10

Could you help please​

Mathematics
1 answer:
Margaret [11]2 years ago
5 0
The 18th term would be 70
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!!!!!!!50 points!!!!!!!! QUESTION BELLOW HELP ASAP!!!!!!!!!
kirza4 [7]
Hmm, interesting
one way would be to multply it out or set it equal to 11x where x is a whole number (if x is not a whole number, then it is not divisible)
11x=7^6+7^5-7^4
undistribute 7^4
11x=(7^4)(7^2+7^1-1)
11x=(7^4)(49+7-1)
11x=(7^4)(55)

56=5*11

11x=(7^4)(5)(11)
divide by 11
x=5(7^4)



aka, find if 11 is a factor of that number


x=5(7^4)
4 0
2 years ago
Read 2 more answers
What is the simplified form of (4p-2)(p-4)​
VikaD [51]

The simplified form is: =4p2−18p+8

Step by Step:

(4p−2)(p−4)

=(4p+−2)(p+−4)

=(4p)(p)+(4p)(−4)+(−2)(p)+(−2)(−4)

=4p2−16p−2p+8

=4p2−18p+8

3 0
3 years ago
HELP ME PLEASE ITS THE LAST QUESTION ON THIS TEST
tensa zangetsu [6.8K]

Answer:

a

Step-by-step explanation:

5 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
Find the area of the figure.
Anettt [7]

Answer: 28.5 square units.

Step-by-step explanation: Separate the figure into a rectangle and a triangle. Count the length and width of the rectangle. The length of the rectangle is 8 units and the width is 3 units. To find the area use the formula l*w. 8*3=24.

Next find the area of the triangle section. The triangle is 3 units tall and 3 units wide. To find the area use the formula 1/2(l*w). 3*3=9. 9/2=4.5.

Finally add the areas of the rectangular section and the triangular section. 24+4.5=28.5.

4 0
3 years ago
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