Fraction as a mixed number in simplest form

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^

guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

$d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j$

B )

Acceleration a is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j$

Hence , this is the required solution .

5 0

2 years ago

Look closely at the model.

Maslowich

Answer:

Perimeter: 2(x+3) + 2(x+2)

Step-by-step explanation:

6 0

2 years ago

HELP I NEED HELP ASAP

Tpy6a [65]

Answer:

I think it's -14

Step-by-step explanation:

x = 0 so the x won't be included but

-2 • x = -2x

-12 • x = -12x

-2x + (-)12x = -14

hope this helps :)

7 0

2 years ago

Evaluate the integral of the quotient of the cosine of x and the square root of the quantity 1 plus sine x, dx.

VMariaS [17]

Answer:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c.

Step-by-step explanation:

In order to solve this question, it is important to notice that the derivative of the expression (1 + sin(x)) is present in the numerator, which is cos(x). This means that the question can be solved using the u-substitution method.

Let u = 1 + sin(x).

This means du/dx = cos(x). This implies dx = du/cos(x).

Substitute u = 1 + sin(x) and dx = du/cos(x) in the integral.

∫((cos(x)*dx)/(√(1+sin(x)))) = ∫((cos(x)*du)/(cos(x)*√(u))) = ∫((du)/(√(u)))

= ∫(u^(-1/2) * du). Integrating:

(u^(-1/2+1))/(-1/2+1) + c = (u^(1/2))/(1/2) + c = 2u^(1/2) + c = 2√u + c.

Put u = 1 + sin(x). Therefore, 2√(1 + sin(x)) + c. Therefore:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c!!!

4 0

2 years ago

Read 2 more answers

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

a)

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

b)

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$