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3 years ago

How do you find the average velocity over an interval?

Mathematics

1 answer:

Artist 52 [7]3 years ago

3 0

If the moving object is undergoing constant acceleration, then

$v_{\rm avg}=\dfrac{v_i+v_f}2$

(take the average of the initial and final velocities, if they're given)

If you don't know the acceleration or initial/final velocities, then you have to resort to the usual definition,

$v_{\rm avg}=\dfrac{\Delta x}{\Delta t}$

where $\Delta x$ is the change in position and $\Delta t$ is the duration of the interval.

You might be interested in

What is the range of the function y = x 2?

garik1379 [7]

Answer:

The range of the function y = x2 is all real numbers, y, such that y ≥ 0. In the case of the function y = x2, the domain consists of all real numbers, because there is no number that, when plugged in for x into x2, would create an undefined expression.

Step-by-step explanation:

7 0

3 years ago

What is the factorization of 3x2 – 8x + 5?

ELEN [110]

we have

$3x^{2} -8x+5$

Equate the expression to zero to find the roots

$3x^{2} -8x+5=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$3x^{2} -8x=-5$

Factor the leading coefficient

$3(x^{2} -(8x/3))=-5$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$3(x^{2} -(8x/3)+(16/9))=-5+(16/3)$

$3(x^{2} -(8x/3)+(16/9))=(1/3)$

Rewrite as perfect squares

$3(x-(4x/3))^{2}=(1/3)$

$(x-(4x/3))^{2}=(1/9)$

square roots both sides

$(x-\frac{4}{3})=(+/-) \sqrt{\frac{1}{9}}$

$x=\frac{4}{3}(+/-) \frac{1}{3}$

the roots are

$x=\frac{4}{3}+ \frac{1}{3}=\frac{5}{3}$

$x=\frac{4}{3}- \frac{1}{3}=\frac{3}{3}=1$

$3x^{2} -8x+5=3(x-\frac{5}{3})(x-1)=(3x-5)(x-1)$

therefore

<u>the answer is the option</u>

$(3x-5)(x-1)$

6 0

3 years ago

Read 2 more answers

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

3 years ago

I dont really understand this so can someone help me please!

pychu [463]

The area of a circle is pi times the radius squared. the radius of this circle is 16.8cm, which is 282cm when squared. multiplied by pi it becomes 887cm.

3 0

3 years ago

Which of the following is the solution to 7√22 + 3√7

Olegator [25]

The value of this irrational Expression is 40.77 .

<h3>What is an Expression ?</h3>

An expression is a mathematical statement consisting of variables , constants and mathematical operators.

The expression given is

7√22 +3√7

√22 = 4.69

√7 = 2.65

32.83 + 7.94

40.77

The value of this irrational Expression is 40.77 .

To know more about Expression

brainly.com/question/14083225

#SPJ1

5 0

2 years ago