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3 years ago

Anyone can help me? please

Mathematics

1 answer:

postnew [5]3 years ago

4 0

If it has a line on top of the last number then it is considered repeating.

therefore, the first two are repeating and the last two are non-repeating.

Hope that helps!

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Find the length of the following two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 < t < 16

andrezito [222]

Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

$r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt$

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

$r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt$

Doing the operations inside of the brackets the derivatives are:

1 ) $(\frac{d((1/2)t^2)}{dt} )^2= t^2$

2) $\frac{(d(1/3)(2t+1)^{3/2})}{dt} )^2=2t+1$

Entering these values of the integral is

$r(t)= \int\limits^{16}_{0} \sqrt{t^2 +2t+1} dt$

It is possible to factorize the quadratic function and the integral can reduced as,

$r(t)= \int\limits^{16}_{0} (t+1) dt= \frac{t^2}{2} + t$

Thus, evaluate from 0 to 16

$\frac{16^2}{2} + 16$

The value is $r= 144 units$

5 0

4 years ago

Help ASAP please!

Iteru [2.4K]

Answer:A

Step-by-step explanation:

it is A

6 0

3 years ago

Read 2 more answers

In circle D, angle ADC measures (7x + 2)°. Arc AC

Akimi4 [234]

Answer:

The measure of angle ABC is 36° ⇒ 1st answer

Step-by-step explanation:

Let us revise some important facts in the circle

The measure of the center angle is equal to the measure of its subtended arc

The measure of the inscribed angle is equal to half the measure of the central angle subtended by the same arc

The vertex of a central angle is the center of the circle and its sides are radii in the circle

The vertex of an inscribed angle is a point on the circle, and its sides are chords in the circle

In circle D

∵ D is the center of the circle

∵ A and C lie on the circle

- DA and DC are radii

∴ ∠ADC is a central angle subtended by arc AC

∴ m∠ADC = m of arc AC

∵ m∠ADC = (7x + 2)°

∵ m of arc AC = (8x - 8)°

- Equate them to find x

∴ 8x - 8 = 7x + 2

- subtract 7x from both sides

∴ x - 8 = 2

- Add 8 to both sides

∴ x = 10

Substitute the value of x in the measure of ∠ADC

∵ m∠ADC = 7(10) + 2 = 70 + 2

∴ m∠ADC = 72°

∵ AB and BC are two chords in circle D

∴ ∠ABC is an inscribed angle subtended by arc AC

∵ ∠ADC is a central angle subtended by arc AC

- By using the 2nd fact above

∴ m∠ABC = m∠ADC

∴ m∠ABC = × 72

∴ m∠ABC = 36°

7 0

3 years ago

Draw the graph for the equation by first finding at least two points on the line.<br> 2x-y=9

ELEN [110]

Answer:

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4 0

1 year ago

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The rang

VARVARA [1.3K]

Answer:

$\theta=24.4^\circ$

Step-by-step explanation:

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$ , where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$ since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$

3 0

3 years ago