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wel
3 years ago
7

Answer these two questions ​

Mathematics
2 answers:
kirill [66]3 years ago
5 0
There is not a question!
sveta [45]3 years ago
4 0

Answer:

Answered

Step-by-step explanation:

You might be interested in
Which point on the scatter plot is an outlier? A scatter plot is shown. Point D is located at 1 and 1, Point C is located at 2 a
katovenus [111]

Answer:

The outlier is:

Point A.

Step-by-step explanation:

We know that an outlier is a point that stands out as compared to the rest of the data points.

Clearly from the scatter plot based on the given points in the question we get tat Point A is an outlier because it attains a different value as compared to the rest of the data points.

Hence, the outlier is:

Point A.

6 0
3 years ago
Read 2 more answers
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
the graph of the absolute value parent function is vertically stretched by a factor of three, then shifted two units left &
ahrayia [7]

Answer:

  • y =  3 | x + 2 | - 5

Step-by-step explanation:

<u>The absolute value function is:</u>

  • y = | x |

<u>Step 1. Function is vertically stretched by a factor of 3:</u>

  • y = 3 * | x |

<u>Step 2. Shifted two units left:</u>

  • y = 3 | x + 2 |

<u>Step 3. Shifted 5 units down:</u>

  • y =  3 | x + 2 | - 5

<em>See attached</em>

<em>Steps are </em><em>Black → Blue → Green → Red</em>

6 0
3 years ago
A jar contains only​ pennies, nickels,​ dimes, and quarters. There are 23 ​pennies, 22 ​dimes, and 20 quarters. The rest of the
goldfiish [28.3K]

<em>Answer:</em>

<em>1. 65 coins</em>

<em>2. 79 - (23 + 22 + 20) = n</em>

<em>Step-by-step explanation:</em>

<em>You can use the expression 23 + 22 + 20 to get the amount of coins that are not nickles. And for two? Try it yourself.</em>

<em>Hope this helps. Have a nice day.</em>

8 0
3 years ago
The sides of a triangle are x , x +1 , 2 x -1 and its area is x root of 10 .Find the value of x.
VARVARA [1.3K]


Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.

Substitute the known values into the formula:

x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}

Simplify:

<span>x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}</span>

<span>x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]</span>

<span>x√10=√[2x(x)(x-1)(1)]</span>

<span>x√10=√[2x²(x-1)]</span>

<span>x√10=√(2x³-2x²)</span>

<span>10x²=2x³-2x²</span>

<span>2x³-12x²=0</span>

<span>2x²(x-6)=0</span>

<span>2x²=0 or x-6=0</span>

<span>x=0 or x=6</span>

<span>Therefore, x=6 (you can't have a length of 0).</span>

6 0
3 years ago
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