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3 years ago

Find the work done by the force field f in moving an object from a to

1 answer:

6 0

Presumably the path from $a$ to $b$ is a line segment, in which case we can parameterize it by

$\mathbf r(t)=(1-t)(1,1)+t(4,9)=(1+3t,1+8t)$

with $0\le t\le1$ . Then the work done along this path - denoted by $\mathcal C$ - is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}(6(1+8t)^{3/2},9(1+3t)(1+8t))\cdot(3,8)\,\mathrm dt$
$=\displaystyle\int_0^1(72(1+3t)(1+8t)+18(1+8t)^{3/2})\,\mathrm dt$
$=\dfrac{6309}5$

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Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

8 0

1 year ago

Can someone pls help me Pls

lora16 [44]

Answer:

help u with what? Say what u need

5 0

2 years ago

Neeeeed helpppppp please

MissTica

Answer:

Scale factor = 7

Step-by-step explanation:

Dilation with scale factor to map HEFG to DABC will be,

Scale factor = $\frac{\text{Dimension of Image DABC}}{\text{Dimension of preimage HEFG}}$

= $\frac{\text{Length of CD}}{\text{Length of GH}}$

Length of CD = Distance between two points C(0, -7) and D(-7, 0)

= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

= $\sqrt{(-7-0)^2+(0+7)^2}$

= $\sqrt{98}$

= $7\sqrt{2}$

Length of GH = Distance between G(0, -1) and H(-1, 0)

= $\sqrt{(-1-0)^2+(0+1)^2}$

= $\sqrt{2}$

Scale factor = $\frac{7\sqrt{2} }{\sqrt{2} }$

= 7

6 0

3 years ago

1. Identify the DEPENDENT VARIABLE in the testable

MAVERICK [17]

To identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

In a research study, there are typically two main variables that direct the scientific enquiry. They are:

Dependent Variable, and
Independent Variable

The independent variable causes a change in the dependent variable, i.e. the dependent variable receives the effect, the independent variable is the cause of the change.

It is very easy to identify the dependent variable in any testable hypothesis once you are able to pick out which variable is causing a change in the other.

For example, let's say the topic of a research is: The Impact of Sunlight on Germination Rate of Seedlings.

Here, Sunlight is the independent variable affecting Germination Rate.

The dependent variable here would be: Germination Rate.

Therefore, to identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

Learn more here:

brainly.com/question/24657192

5 0

2 years ago

How to find the median

KengaRu [80]

We can find the median by using the formula:

(N+1/2) th item

For example:

Given data: 4,6,3,5,7

We have to arrange the data into ascending order:

=3,4,5,6,7

N(total no of items)=5

Now,

Median=(N+1/2)th item

 =(5+1/2) th item

 =(6/2) th item

 = 3rd item

Median=5

Hope it helps

Good luck on your assignment

3 0

3 years ago

Read 2 more answers