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4 years ago

Given the points (-1,-5) and D(-7, 11). find the coordinates of point E

Mathematics

1 answer:

valkas [14]4 years ago

3 0

Answer:

The coordinates of point E are (-19/4,8)

Step-by-step explanation:

To find this, we use the internal division formula;

The formula is as follows;

(x ,y) = mx2 + nx1/(m + n) , my2 + ny1/(m + n)

From the question;

m = 5 n = 3

x1 = -1 , y1 = -5

x2 = -7 , y2 = 11

Substitute these values;

(x ,y) = 5(-7) + 3(-1)/(3 + 5) , 5(11) + 3(-5)/(5 + 3)

(x , y) = (-35-3)/8, (55-15)/8

(x,y ) = -38/8, 40/8

(x,y) = -19/4, 5

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4 years ago

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Your truck has a weight limit of 2100 pounds. A bag of cement weighs 75 pounds. Write an equality for x, the number of bags of c

Hitman42 [59]

You can haul 28 bags of cement.

Step-by-step explanation:

Given,

Weight limit of truck = 2100 pounds

Weight of each bag of cement = 75 pounds

Let,

x be the number of cement bags that you can haul

Weight of each bag * Number of bags = Weight limit

$75x = 2100$

Dividing both sides by 75

$\frac{75x}{75}= \frac{2100}{75}\\x=28$

You can haul 28 bags of cement.

Keywords: division, equality

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#LearnwithBrainly

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3 years ago

A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha

Arlecino [84]

Answer:

A) $F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }$ with $a = \frac{2\Delta x}{t^2}$

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

$\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha)$ (1)

where $\alpha$ is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration $a$ that can be calculated as a function of the given displacement and time interval as:

$a = \frac{2\Delta x}{t^2}$ and the force equation is:

$\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma$

now substituting $N$ from equation (1):

$F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}$