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3 years ago

Which of the following statements about 585 are true? (Choose 2)

Mathematics

2 answers:

wariber [46]3 years ago

8 0

Answer:

Statements about true about 585 are:

B. it is divisible by 3

C. it is divisible by 5

Step-by-step explanation:

Lets find out the all factors of 585 because all the factors of 585 are the perfect dividers of 585.

So first we will find out the Lowest Common Factors or LCM of 585

585=1×3×3×5×13

Hence , the perfect dividers of 585 are 1,3,5,13,

That means 3, 5 can be the dividers of 585.

Natali5045456 [20]3 years ago

5 0

5 and 3 are because with 5 it is 117 and 3 it 195.

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The difference between -7 and a number x is no more than -2/3

Troyanec [42]

Answer:

Pregunta muy difícil, no creo que pueda resolver este amigo.

Step-by-step explanation:there is no step-by-step thing because honestly don't understand what your trying to say here. Hmm...

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

3 years ago

A circle has a radius of 2.5 cm. What is the diameter of this circle in cm? Record your answer in Blank 1. What is the circumfer

Eva8 [605]

Answer:

The diameter is 5 cm and the circumference is diameter x pi

Step-by-step explanation:

3 0

3 years ago

A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation

kirill [66]

Answer:

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.25

The margin of error is:

M = T*s = 3.25*25 = 81

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds

The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

6 0

3 years ago

Cranberry juice costs $6.30 per quart and apple juice costs $3.60 per quart. Terrence wants to know how many quarts q of cranber

Korvikt [17]

We know that:

Cost (Cranberry Juice): 6.3 per quart

Cost (Apple Juice): 3.6 per quart

Quantity (Apple Juice): 4 quarts

Quantity (Cranberry Juice): q quarts

Further we know that Terrence wants the cost of the juice to be 4.5 per quart

Hence Total Cost = Cost per quart × Total Number of quarts = (Cost per quart for Apple Juice × Total Number of Apple Juice Quarts) + (Cost per quart for Cranberry Juice × Total Number of Cranberry Juice Quarts)

⇒ Total Cost = 4.5 × (q+4) = 6.3 × q + 3.6 × 4

⇒ 4.5(q+4) = 6.3q + 14.4, which matches with option D

Hence, the correct option is D

6 0

3 years ago

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