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2 years ago

Please help I need to pass

Mathematics

2 answers:

Julli [10]2 years ago

5 0

I think the answer is

Archy [21]2 years ago

5 0

Answer:

hope this answer helps you dear...take care and may u have a great day ahead dear!

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The surface area of the solid is about ____ inches squared

Dafna11 [192]

Answer:

Surface area of the whole figure $= 320.28inch^2$

Step-by-step explanation:

The surface area of the figure :

Surface area of the cylinder(excluding top)+Surface area of the cone(excluding bottom)

Radius of the cylinder=3 inch and Height of cylinder=13 inch

Surface area of the cylinder with one circular face

$=2\pi rh+\pi r^2$

As, $\pi =\frac{22}{7}= 3.14$

$=(2*3.14*3*13)+(3.14*3*3)\\\\=244.92+28.26\\\\=273.18inch^2$

Surface area of the cone excluding base:

$=\pi *r*l$

l= slant height=5 inch r= radius= 3 inch

Surface area of the cone = $3.14*3*5$

$=47.1inch^2$

Total surface area of the whole figure $=273.18+47.1$

$= 320.28inch^2$

5 0

3 years ago

The guidance department has reported that of the senior class, 2.3% are members of key club, K, 8.6% are enrolled in AP physics,

Levart [38]

P(P|K) = 82.6%.

P(P|K) = P(K and P)/P(K) = 1.9%/2.3% = 0.019/0.023 = 0.8261 = 82.6%

4 0

3 years ago

Read 2 more answers

Solving an Equation for a Given Variable

Reil [10]

Answer:

$a=\frac {f-v}{t}$

Step-by-step explanation:

f=v+at

It requires you to make a the subject of the formula

f-v=at

Dividing both sides by t then

$a=\frac {f-v}{t}$

Therefore, the equivalent for a is $a=\frac {f-v}{t}$

8 0

3 years ago

Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o

umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)

1/2 ∫▒(2xdx/u) =1/2 ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2 ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C (required soultion)

3 0

3 years ago

Read 2 more answers

Factor LaTeX: 7v^2+31v+12

bazaltina [42]

Answer:

(7v +3 ) ( v + 4 )

Step-by-step explanation:

7v^2+31v+12

(7v + ) ( v + )

12 = 1*12

2*6

3*4

7* 4 + 3 = 31

(7v +3 ) ( v + 4 )

6 0

3 years ago