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3 years ago

The robots are powered on coffee alone. They use 2 bags of coffee each day. How long will 1/4 of a bag of coffee last them

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

8 0

probably about 1//8th of a day

Vadim26 [7]3 years ago

5 0

2 bags = 24 hours
Therefore 1 bag = 12 hours, half a bag = 6 hours and 1/4 of a bag must be? 3 hours :)

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What is the solution to the equation 3/m+3-m/3-m=m^2+9/m^2-9

Bas_tet [7]

Answer:

no solution

Step-by-step explanation:

If we subtract the left side of the equation, we get ...

$\dfrac{m^2+9}{m^2-9}-\dfrac{3}{m+3}+\dfrac{3}{3-m}=0\\\\\dfrac{m^2+9-3(m-3)-3(m+3)}{(m-3)(m+3)}=0\\\\\dfrac{m^2-6m+9}{(m-3)(m+3)}=0=\dfrac{(m-3)^2}{(m-3)(m+3)}\\\\\dfrac{m-3}{m+3}=0\quad\text{m$\ne$3}$

This equation will equal zero only if m=3, which is disallowed because it makes the denominator zero. Thus, there is no solution.

7 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

at a track meet,Jacob and daniel compete in 220 m hurdles. Daniels finishes in 3/4 of the a min.Jacob finishes with 5/12 of a mi

Savatey [412]

Jacob has the faster time because if you make the fractions equal you would make the denominators equal by rounding 3/4 to have a denominator of 12 making it 9/12 which is larger then 5/12 making Daniel's time slower then Jacob's.

3 0

3 years ago

Heather won 100 lollipops playing horseshoes at her school's game night. Later, she gave four to each of her friends. She only h

vova2212 [387]

Answer:

23 friends

Step-by-step explanation:

100 - 8= 92

92/4 = 23

HEATHER IS SO POPULAR!!

8 0

3 years ago

What is the rate of decay for the function f(x) =2(0.95)^x ( x is on top of the parentheses)

SIZIF [17.4K]

Answer:

The decay rate is 5%.

Step-by-step explanation:

Let a substance is decaying at the rate of r% per hour from the initial value of P for t hours, then the final value of the substance is given by the function

$f(t) = P(1 - \frac{r}{100})^{t}$ ........... (1)

Comparing this equation with the original equation given as

$f(x) = 2(0.95)^{x}$ ............ (2) we get,

$(1 - \frac{r}{100}) = 0.95$

⇒ $\frac{r}{100} = 0.05$

⇒ r = 5%.

Therefore, the decay rate is 5%. (Answer)

7 0

3 years ago