All categories

2 years ago

Andrew earned $125.00 last year on his investment of $10,000.00

Mathematics

2 answers:

loris [4]2 years ago

8 0

Answer:

$500

Step-by-step explanation:

Artemon [7]2 years ago

7 0

Answer:

1,25%

Step-by-step explanation:

Aby policzyć stopę procentową należy wykorzystać podzielić otrzymany zarobek (125$) przez kwotę zainwestowaną (10.000$) a następnie pomnożyć przez 100, aby otrzymać wartość wyrażoną w procentach.

I tak mamy

$\frac{125}{10000}$ x 100 [%] = 1,25%

You might be interested in

Lisa’s cameras has 2,048 megabytes of memory for storing pictures. She already used half this amount. A high-quality pictures us

Gala2k [10]

Answer:

Step-by-step explanation:

divide 2048 by 2 since half is already used gives you 1024 ...then divide 1024 by 16 to get 64..

4 0

2 years ago

1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the

Kryger [21]

<span>1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2

D.y=-x

2.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-1);y=-3/2x-6

C.y=-3/2x+2

3.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (4,2);x=-3
D.y=4

4.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (-2,3);y=1/2x-1

B.y=-2x-1

5.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (5,0);y+1=2(x-3)

D.y=-1/2x+5/2</span>

5 0

2 years ago

Read 2 more answers

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

The histogram below shows information about the number of books students read in a month:

max2010maxim [7]

Answer:

Option C is true - The number of students who read 4 books or fewer

Step-by-step explanation:

Here, to find the mean and median, we do not have proper information or specific values. We just have the ranges, so both the mean and median cannot be determined. So, options A and B are not correct.

We can see two bins covering the 9 or more books, so option D cannot be true either.

Therefore, only option C is left and that is true.- the number of students who read 4 books or fewer.

6 0

2 years ago

-20 + 11= Please help if you know what the answer is please show work or explane how you got it?

Travka [436]

Step-by-step explanation:

-20+11

negative and positive sign will change in negative so ans will be=-9

because positive sign is on greater number

6 0

2 years ago

Read 2 more answers