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PilotLPTM [1.2K]
3 years ago
14

PLz answer before 11:59 pm

Mathematics
1 answer:
Tpy6a [65]3 years ago
6 0

Answer:

SA = 2520 ft^2

Step-by-step explanation:

a = 21ft

b = 29ft

c = 20ft

h = 30ft

SA = 2A_{B} + (a+b+c)h

A_{B} = s(s﹣a)(s﹣b)(s﹣c)

s = \frac{a + b + c}{2}

SA= ah + bh + ch + \frac{1}{2} \sqrt{-a^4 + 2(ab)^2 + 2(ac)^2 - b^4 + 2(bc)^2 - c4}

SA= (21 * 30) + (29 * 30) + (20 * 30) +

       \frac{1}{2} \sqrt{-(21)^4 + 2(21 * 29)^2 + 2(21 * 20)^2 - (29)^4 + 2(29 * 20)^2 - (20)^4}

SA = 2520 ft^2

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Do not give me any links or files.
Paha777 [63]

Answer:

972 u³

Step-by-step explanation:

The radius is 9 . We can use formula ,

⇒ V = 4/3 π r³

⇒ V = 4/3 × π × 9³

⇒ V = 4 × π × 81 × 3

⇒ V = 972 unit³

7 0
3 years ago
2:14 = 1:<br> What is the equivalent ratio?
Vlad [161]
The answer would be 1:7
7 0
3 years ago
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2x2 − 1 = 3x + 4 Which equation correctly applies the quadratic formula?
NNADVOKAT [17]

Standard form: ax^2 + bx + c = y

Steps to simplify:

2x^2 - 1 = 3x + 4

~Subtract (3x + 4) to both sides

2x^2 - 1 - 3x + 4 = 3x + 4 - 3x + 4

~Simplify

2x^2 - 3x - 5 = 0

(a = 2, b = -3, c = -5)

Now that we know what these values are, we can use the quadratic formula to solve.

x=\frac{-b-+\sqrt{b^2-4ac} }{2a} \\x = \frac{-(-3)-+\sqrt{-3^2-4(2)(-5)} }{2(2)}\\x= \frac{-3-+\sqrt{49} }{4} \\x=\frac{3-+7}{4} \\x = \frac{10}{4}~or~\frac{-4}{4}  \\x=\frac{5}{2}~or~-1

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Best of Luck!

5 0
3 years ago
Which is true about rigid transformations and circles?
steposvetlana [31]

A rigid transformation is defined to be one in which the pre-image of the object and its new image after the transformation both have the exact same size and shape. So the answer in this question is:

<span>D. a circle's shape is preserved regardless of any rigid transformation</span>

5 0
3 years ago
The function a(t)=t^(1/2)−t^(−1/2) m/s^2 represents the acceleration of a particle moving along a horizontal axis. At time t=0,
garri49 [273]

Answer:

see below

Step-by-step explanation:

a(t)=t^(1/2)−t^(−1/2)

We integrate to find the velocity

v(t) = integral t^(1/2)−t^(−1/2) dt

     = t ^ (1/2 +1)         t ^ (-1/2 +1)

          ------------   -    -----------------  + c  where c is the constant of integration

              3/2                   1/2

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +c

We find c by letting t=0 since we know the velocity is 4/3 when t=0

v(0) = 2/3 0^ 3/2  - 2 0^ 1/2 +c = 4/3

       0+c =4/3

       c = 4/3

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +4/3

To find the position function we need to integrate the velocity

p(t) = integral 2/3 t^ 3/2  - 2 t^ 1/2 +4/3 dt

     2/3 t ^ (3/2 +1)        2 t ^ (1/2 +1)           4/3t

          ------------   -    -----------------  + ------------- + c  

              5/2                   3/2                    1

p(t) =  4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t +c

We find c by letting t=0 since we know the position is -4/15 when t=0

p(0) =  4/15 0^ 5/2 - 4/3 0 ^ 3/2 + 4/3*0 +c = -4/15

         0 +c = -4/15

            c = -4/15

p(t) =  4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t -4/15

8 0
3 years ago
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