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Vanyuwa [196]
3 years ago
9

Which expression is equivalent to 4.2a−2(0.4a−1.3)

Mathematics
2 answers:
Serggg [28]3 years ago
8 0
Is the answer 4a+1.3
Sati [7]3 years ago
3 0

Answer:

the 3rd one

Step-by-step explanation:

You might be interested in
Neggative 4 greater or smaller than neggative 5
Vanyuwa [196]

Answer

Smaller

Step-by-step explanation:

The closer it is the 0 the smaller

5 0
3 years ago
I don’t know how to do it
mote1985 [20]
Basically, we need to find the diagonal of the square. Once we have the diagonal, we need to find if it's more or less than 11 cm, which is diameter of the circle.
This is the formula we'll use:
a²+b²=c²
Since it's a square, a and b are the same.
7²+7²=c²
49+49=c²
98=c²
√98=c
c=9.8995
So, the diagonal is 9.8995
9.8995<11

So, since the diagonal of the square is 9.8995 cm, it is less than the circumference of the circle meaning that it will fit in to the circle without touch the circle's circumference.
6 0
3 years ago
1 Sophia is making a mosaic pattern with similar
LuckyWell [14K]

Answer:

C.) Rectangle JKLM

Step-by-step explanation:

Multiply all dimensions in Rectangle JKLM by 2 and they will be equal to the dimesions of Rectangle ABCD

3 0
3 years ago
What is the length of PR
goldenfox [79]

Answer:

7

Step-by-step explanation:

the ratio is 3 :1

here is your answer

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
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