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3 years ago

Kyle has a lock combination with three

Mathematics

2 answers:

Tpy6a [65]3 years ago

8 0

Answer:

kYL E DID WHAT

Step-by-step explanation:

kaheart [24]3 years ago

7 0

Answer:

Step-by-step explanation:

There are 5 * 5 * 5 = 125 possible combinations. (Duplications are allowed, otherwise you wouldn't be asking about 000. )

There are 5 choices for each number in the lock. There is only 1 way that you can get 0 0 0

The probability is 1/125 or 0.008

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Which equations represent the data in the table? Check

Studentka2010 [4]

Answer:

A,D, and E are the answers on E2020

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3 years ago

A bank is advertising the following savings account.

Vlad1618 [11]

Answer:

6.66%

Step-by-step explanation:

((1 + r/m)^n) - 1

Interest rate , r = 6.5%

m = number of compounding times per period = 4 (quarterly interest)

((1 + 0.065/4)^4) - 1

((1 + 0.01625) ^4) - 1

((1.01625)^4) - 1

1.06660 - 1

= 0.0666

0.0666 * 100%

= 6.66%

8 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 $0.31\; \rm m$ .

7 0

2 years ago

Of all the students enrolled at the American dream, 9% are playing a music instrument. There are 330 students in ADS. How many s

Genrish500 [490]

Exactly 29.7, but you can't have 7/10 of a student so you round to 30. So 30 students are playing a music instrument because 330 times 0.09 is about 30

5 0

3 years ago

If three points are coplanar, they are also collinear true or false

Darina [25.2K]

I am sorry i don’t understand can you just type it again ?

5 0

3 years ago

Read 2 more answers