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Please provide steps.
Solve number 56.

2

Step-by-step explanation:

∫₀⁴ p / √(9 + p²) dp

If u = 9 + p², then du = 2p dp.  So ½ du = p dp.

When p = 0, u = 9.  When p = 4, u = 25.

∫₉²⁵ (½ du) / √u

½ ∫₉²⁵ u^-½ du

½ (2u^½) |₉²⁵

√u |₉²⁵

√25 − √9

5 − 3

2

2

Step-by-step explanation:

We will use a u substitution

Let u = 9+p^2

du = 2p dp

The lower limit becomes

u = 9+0^2 = 9

The upper limit becomes

u = 9+4^2 = 25

   where a is 25

We know the intergral of 1 / sqrt(u)  is  sqrt(u)/ 1/2

1/2 * sqrt(u)/(1/2) evaluated at 25 and 9

sqrt(u)  where u is25 - sqrt(u)  where u is 9

sqrt(25) - sqrt(9)

5-3

2

i need more information

step-by-step explanation:

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm