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ankoles [38]
2 years ago
6

Can you answer this quick​ it's so hard

Mathematics
2 answers:
bulgar [2K]2 years ago
7 0

1. The shape of the base is a square.

2. The lateral faces are triangles.

3. Including the square and four triangles, the figure has 5 faces.

4. Square pyramid.

5. Find the surface area of the square. Find the surface area of one triangle, and multiply by 4 to get all of them. Add the area of the four triangles to the area of the square.

Hope this helps.

Bingel [31]2 years ago
6 0

Answer:

square

triangle

5

square pyramid

find area of base and area of a lateral face (base is 1 side squared, lateral face is base * height / 2. Multiply area of lateral face by 4 and add the product to the area of the base to find the Surface Area.

Step-by-step explanation:

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A of the figure. (Sides meet at right angles.)
Vlada [557]

Answer:

first one is a of 0,12

Step-by-step explanation:

5 0
3 years ago
There are 7 unique names in a bowl. In how many orders can 2 names be chosen? Hint: The word orders
Lelu [443]

Answer:

2/7

Step-by-step explanation:

Probability=number of names chosen/total number of names

p=2/7

3 0
3 years ago
The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi
nadezda [96]

Answer:

Step-by-step explanation:

We have volume of cone as

V=\frac{1}{3} \pi r^2 h

and for a cone always r/h = constant

Given that r' = rate of change of radius = -7 inches/sec

(Negative sign because decresing)

V' =- 948 in^3/sec

Radius = 99 inches and volume = 525 inches

Height at this instant = \frac{525}{\frac{1}{3} \pi (99)^2} \\=\frac{0.1607}{\pi}

Let us differentiate the volume equation with respect to t using product rule

V=\frac{1}{3} \pi r^2 h\\V' = \frac{1}{3} \pi[2rhr'+r^2 h']\\-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\

-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\-948 = 33(3.14)(-2.25/3.14  + 99 h')\\-9.149=-0.72+99h'\\-8.429 = 99h'\\h' = 0.08514

Rate of change of height = 0.08514 in/sec

8 0
3 years ago
A mass of 3.25 kg is attached to the end of a spring that is stretched 22 cm by a force of 15 N. It is set in motion with an ini
mylen [45]

Answer:

Step-by-step explanation:

Given that,

Mass of object=3.25kg

The extension e=22cm=0.22m

Force applied to cause extension F=15N

Initial position Xo=0

Initial velocity Vo=-12m/s

We can get the spring constant from Hooke's law

F=ke

Then, k=F/e

k=15/0.22

k=68.182N/m

Also our natural frequency w is given as

w=√(k/m)

Therefore,

w=√(68.182/3.24)

w=√20.98

w=√21

w=4.58rad/s

w=4.6rad/s

There is no damping in this situation, no outside force acting on the system and the equation that governs the system is

mx''+kx=0

3.25x''+68.182x=0

Divide through by 3.25

x''+20.98x=0

We can approximate 20.98 to 21

x"+21x=0

The solution to this differential equation using D operator

D²+21=0

D²=-21

D= ±√-21

D=±√21 •i

Then the solution is

x(t)=A•Sinwt +B•Coswt

x(t)=A•Sin√21 t +B•Cos√21 t

Note that x'(t)=v(t)

and at t=0 Vo=-12m/s

x(t)=A•Sin√21 t +B•Cos√21 t

x'(t)=v(t)=A√21•Cos√21 t - B√21•Sin√21 t

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

Then, using the two initial conditions

v(0)=-12

And X(0)=0

x(t)=A•Sin√21 t +B•Cos√21 t

X(0)=A•Sin√21•0 +B•Cos√21•0

X(0)=A•Sin0+B•Cos0

0=B

B=0

Also,, V(0)=-12m/s

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

V(0)=A√21•Cos√21•0- B√21•Sin√21•0

V(0)=A√21•Cos0- B√21•Sin0

-12=A√21

Therefore,

A=-12/√21

A=-2.62

Therefore the general equation becomes

x(t)=A•Sin√21 t +B•Cos√21 t

x(t)=-2.62Sin√21 t +0•Cos√21 t

x(t)=-2.62Sin√21 t

a. The amplitude

Comparing x(t) to wave equations

x(t)=-Asin(wt+2λ/t)

Then,

A=2.62m

b. We know the natural frequency already to be

w=√21

w=4.58rad/s

c. Period

Comparing the equation again

wt=√21t

Given that w=2πf

Therefore, 2πft=√21t

Then, f=√21t / 2πt

f=√21/2π

f=0.73Hz

Then, period is the reciprocal of frequency

T=1/f

T=1/0.73

T=1.37seconds

The period is 1.37sec,

5 0
3 years ago
Is the answer right for number 14 if you answer I will be your friend and give you easy answers so you can get points plus I wil
Tema [17]
I think so:

Your working out is probably just the same as my working out!
4 0
3 years ago
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