What is 2x+y=12 3x-18=y

Given k(x)=x^2+3x-1 and p(x)-7x+10 <br><br> Find k(4)+p(-2)

skelet666 [1.2K]

K(4) = (4)^2 +3x-1
k(4) = 27
p(-2) = -7(-2)+10
p(-2) = 24
Answer: k(4) + p(-2) = 51

8 0

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):

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

and putting the expression for the acceleration:

$\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}}$ (2)

which is the requested expression

B) Substituting the values in (2) and using $g=9.8\ \rm{ms^{-2}}$ :

$F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}$

The gravitational force acting on the box is:

$F_g=mg=50\cdot9.8=490\ \rm{N}$

$F/F_g = (117.634/490)\cdot 100 = 24 \%$

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0

4 years ago

In need of some help please.

Hatshy [7]

By the Pythagorean theorem
50^2 = x^2 +(x +34)^2
2500 = 2x^2 +68x + 1156
x^2 +34x -672 = 0
(x -14)(x +48) = 0
x = 14 or -48

The distance from the wall to the base of the ladder is 14 ft.

_____
7-24-25 is a Pythagorean triple. The dimensions here are double those values. It can be handy to know a few of the Pythagorean triples, as that can let you write down the answers to problems without having to go through the equations.

8 0

3 years ago

BRAINLIEST HELP ASAP!!!

Vesna [10]

2 triangles with a base of 8 and height of 6 divide the base and height in half to get the Base and height of 1 triangle and then use 1/2 B•H to find the area of one triangle 1/2 (4•3) = 6

7 0

3 years ago

Square roots can be compared and ordered by comparing and ordering the numbers underneath the radical symbol.

horsena [70]

Answer:

That is correct, if you're asking if it's true or false

Step-by-step explanation:

if u could subscribe to my friend penny_pg3d that would be great!

6 0

3 years ago