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2 years ago

9

Three friends decide to rent an apartment and split the cost evenly. They each paid $710 towards the total move in cost of first

and last month's rent and a security deposit.
If rent is $690 per month, how much was the security deposit?

1 answer:

lina2011 [118]2 years ago

7 0

Answer:

Help me with this please

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Wich is the greatest six digit number 860,553 or 865,530

Serjik [45]

865,530 is the greatest because the other number, 860,530 has '0' in the place of '5' in the 865,530.

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2 years ago

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A manufacturing company has the following fixed monthly costs and unit variable costs: Rent = $2,200 Utilities = $375 Variable c

valentinak56 [21]

A manufacturing company has the following fixed monthly costs and unit variable costs:

Rent = $2,200

Utilities = $375

Variable costs for materials and assembly = $13.25/unit

Monthly labor = $750

Given is the product is sold at $24.99 per unit

Let the number of units sold be x

Adding total cost = $2200+375+750=3325$

Equation becomes:

$3325+13.25x=24.99x$

$11.74x=3325$

x= 283.21

So, rounding it to nearest whole number we get, 283 units.

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2 years ago

What is the measure of angle SQR?

Mademuasel [1]

Answer:

4x°= 4x18= 72°=measure of angle SQR

Step-by-step explanation:

6x +4x = 180

10x = 180

x = 180/10

x= 18°

because m L SQR =4x°

now place the value of x in the above equation

m L SQR = 4 x(18°)= 72°

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3 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

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2 years ago

X two the 2 power =144

Alina [70]

The answer to your question is 12

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