Ask question

Ask question

All categories

3 years ago

7

Find the surface area of the prism formed by the net

1 answer:

Lunna [17]3 years ago

3 0

ANSWER

$T.S.A=1960m^2$

EXPLANATION

The diagram represents the net of a rectangular prism.

The dimensions are,

$l=42m$

$w = 14m$

and

$h = 7m$

The total surface area of a rectangular prism is given by:

$T.S.A=2(lw + lh + wh)$

We substitute the values to obtain,

$T.S.A=2(42 \times 14 + 42 \times 7 + 14 \times 7)$

$T.S.A=2(980)m^2$

$T.S.A=1960m^2$

You might be interested in

Pa help po pls pls pls

Llana [10]

Answer:

acute angle and number 3 is obtuse

Step-by-step explanation:

6 0

2 years ago

Simplify<br> 4^9/4^3<br> A)4^5<br> B)4^6<br> C)4^12<br> D)4^27

atroni [7]

$\frac{4^9}{4^3}=4^{9-3}=4^6$

B)4^6

7 0

3 years ago

Natalie paid $3.84 in sales tax on an item that cost $44.00 before tax. At that rate, how much would she pay in sales tax for an

balandron [24]

Answer:

$70.84

Step-by-step explanation:

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

P(4)+5>-32<br> Thank you ;)

sweet-ann [11.9K]

Answer:

Inequality- 4p + 5 >-32

Step-by-step explanation:

5 0

3 years ago