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1 year ago

5

Kyle took a total of 9 pages of notes during a 3hr class. How many hours of class will it take for Kyle to take 12 pages of ntoe

s?

1 answer:

Vladimir [108]1 year ago

5 0

Answer:

4 hours

Step-by-step explanation:

9÷3=1

1 hour for 3 pages

12÷3= 4

4 hours

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On Tuesday, the Madison Avenue Hotel had 5 occupied rooms for every 2 unoccupied rooms. If there

Step2247 [10]

Answer: say what need more information

Step-by-step explanation:

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3 years ago

Which equation can be solved by using this expression?

Marat540 [252]

<u>Answer:</u>

The correct answer option is B. 2 = 3x + 10x^2

<u>Step-by-step explanation:</u>

We are to determine whether which of the given equations in the answer options can be solved using the following expression:

$x=\frac{-3 \pm\sqrt{(3)^2+4(10)(2)} }{2(10)}$

Here, $a = 10, b = 3$ and $c=-2$ .

These requirements are fulfilled by the equation 4 which is:

$12=3x+10x^2$

Rearranging it to get:

$10x^2+3x-2=0$

Substituting these values of $a,b,c$ in the quadratic formula:

$x= \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

$x= \frac{-3 \pm\sqrt{(3)^2-4(10)(-2)} }{y}$

3 0

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

A soap factory randomly inspects 100 bars of soap to better identify problems that occur during the manufacturing

Mekhanik [1.2K]

Answer:

idk

Step-by-step explanation:

sorry

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3 years ago

Read 2 more answers

An engineer is standing 250 meters away from a building. She measures the angle of elevation to the top of the building to be 30

RSB [31]

Answer:

144m

Step-by-step explanation:

we would used tan 30 to determine the height of the building

tan 30 = opposite / adjacent

0.5774 x 250 = 144m

3 0

3 years ago