Need HELP ASAP PLEASEEEEEEE

Is a dilation an example of a rigid motion. Explain your response.

Likurg_2 [28]

Answer:

This differs from non-rigid motion, like a dilation, where the size of the object can increase or decrease. When working with rigid motion, you will typically see two objects that show the transformation.

Step-by-step explanation:

5 0

2 years ago

Give an example of when you had to prove something, and how you did it!

Marizza181 [45]

Answer:

Just practice ,know different ways of solving it ,gain experince ,be smart and skilled.

Step-by-step explanation:

if you want to prove anuthing then go with the concept accoeding to the nature of the problem and there are almost many ways to prove anything so don't waste your time on way get on to the other ways.

It may not get proved at first time so keep on doing questions and when you have gained experience you know every nature of problems and then you are skilled properly.

8 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Jamir is training for a race and is running laps around a field. If the distance around the field is 250 yards, how many complet

aleksley [76]

Answer:

27 laps

Step-by-step explanation:

1 mile = 1,670 yards

1,670 x 4 = 6,680

6,680 / 250 = 26.72 rounded to the nearest hundreth = 27

6 0

3 years ago

PLEASE HELP QUICKLY 15 POINTS, WILL MARK BRAINLIEST

zhuklara [117]

Answer:

Binomial expansion also known as binomial theorem is a quick and easy way of expanding by multiplying binomial expression that has been raised to large power.

Step-by-step explanation:

A binomial is a polynomial with two terms. From the above definition given, if for instance an expression of (5x-4)²² was given, it wouldn't be that easy to multiply out by hand which is why the binomial theorem formula was created.

Another example to buttress this point is if we have (a³+2ab+b2)(a+b) expanding this will give a³+3a2b+b³ = a³+3a2b) +3ab²+b³)a+b = a⁴+ 4a³b+ 6a²b²+4ab³+b4..

This calculation could go on and on because it gets longer as we go but to make it easier and simpler, the best method to use here is the binomial theorem or expansion and the formula states thus:

n. n

(x+y)ⁿ = € {n}xⁿ-k*y^k

k=0 {k} = € {n} x^k*y^n-k

k=0 {k}

{n}

{k}. = binomial coefficient of each term

It should from the example given that we have index power(power raised)called exponent such that when an exponent is 0, we get 1 , i.e anything raised to power 0 =1

e.g (x+y)^0 = 1

And also when an exponent is 1, then the answer remain unchanged such that (x+y)¹= x+y. An exponent of 2 means to multiply given term by itself in two times and so on.

So to expantiate an expression like

a³+3a²b+3ab²+b³.. This expression has its index power(exponent) start with 3 such that .

The a goes down in the form.. 3,2,1,0 while the exponents of b go upwards in the form 0,1,2,3coeffiThe terms 0-n goes thus

k=0 k=1 k=2 k=3

a³ a² a 1

1 b. b² b³

which equals an-kbk.. This is confusing right. This is how it works

k=0 k=1 k=2 k= 3

an-kb^k. an-kb^k an-kb^k an-k^bk

a³-0b^0. a3-1b¹ a3-2b2 a3-3b3

a³ a²b. ab² b³

So when we put all this together we have: a³+a²b+ ab²+ b³. Actually this answer is correct but what we are missing here is the coefficient of n because we needs to include it.

So this becomes a³+3a²b +3ab²+b³ , the coefficient of n here is 3.

Now if we are to look at the result we got before from(a+b)^0 up to (a+b)³

And here that we have 3 as coefficient but where it is not shown we put 1 which is.

1

1a+ 1b

1a+2ab+ 1b²

1a³ +3a²b+3ab²+1b³... We solve this using pascal triangle.

1 This is solved by adding

1 1 each two numbers above

1 2 1 together to get the one

1 3 3 1 below.

1 4 6 4 1

This pascal pattern can be used for any exponents..1,2,3,5....60...120....n

Binomial theorem is basically the essence of this Pascal triangle pattern.

As stated earlier that the formula of binomial theorem is

n

€{n}a^n-kb^k

n-1 where €= summation

n= numbers given

k=coefficient of n

Now, let's to put into practical by solving this expression.

(x+y)ⁿ=€ⁿ {n}a^n-k* b^k

k=0 {k}

But before we solve this, its important we know the basis of Pascal triangle by using factorial to solve it with the formula n!/k!(n-k)! Such that

Factorial is n multiplied in descending order(i.e multiplying Numbers between 1-n)

K- coefficient of n

N- product of all the numbers between 1&n. For instance we are given this 4!/3!(4-3)! = 4!/3!*1!

= 4*3*2*1/3*2*1 *1 = 24/6 = 4

Now let's go back to the previous example of (x+y)ⁿ where n=4

(x+y)⁴= €⁴ {4} x^4-k*y^k

k=0 {k}

(x+y)⁴= €{4} (x^4-0)y^0 + {4} (x^4-1) *y¹

{0} {1}

+{4} (x^4-2)*y² + {4} (x^4-3)*y³

{2} {3}

+ {4} (x^4-4) * (y⁴)

{4}

= 1*x⁴y^0 + 4*x³y¹ + 6*x²y² +4*x¹y³ +1*x^0y⁴=

x⁴+4x³y +6x²y² + 4xy³ +y⁴(Final answer)

Please note that:

When solving binomial theorem expression, the easiest way to do this is by remembering the patterns below:

* first term's exponent start at n and go down

* second term's exponent start at 0 and goes up

* coefficient are solved with pascal triangle using n!/k!(n-k)!

5 0

2 years ago