I need help with math can u help me

F(x)=3x^2+6x-18 , find f(10)

lord [1]

Answer:

f(10)=342

Step-by-step explanation:

f(10)= 3(10)^2+6(10)-18

complete your exponet first and multiply 6(10)

f(10)=3(100)+60-18

multiply the 3(100)

f(10)=300+60-18

add and subtract all the numbers together

f(10)=342

5 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):

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