PLEASEEEEE HURRRRY SOMBODY ANSWER THESE QUESTIONS FOR MEEEE

Julie says 10% of a number will always be less than 20% of any other number complete one any quality to support Julie’s claim an

blsea [12.9K]

10% of 5 is .5

20% of 10 is 2.

Julie is correct.

10% of 100 is 10.

20% of 15 is 3.

Julie is incorrect.

4 0

4 years ago

According to the remainder theorem, what is one factor of the cubic expression x³+3x²+x-5?

Gnom [1K]

Answer:

x - 1

Step-by-step explanation:

To get one factor of the cubic expression x³+3x²+x-5 according to the remainder theorem we can say that

x³+3x²+x-5 = 0

Then we apply Ruffini's Rule and obtain

x³+3x²+x-5 = (x-1)(x²+4x+5)

where x = 1 is the only real root of the polynomial expression.

4 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):

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