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3 years ago

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A garden store bought a fountain at a cost of $995.66 and marked it up 100%. Later on, the store marked it down 25%. What was th

e discount price?

2 answers:

Aliun [14]3 years ago

7 0

Answer:

Step-by-step explanation:

nexus9112 [7]3 years ago

3 0

Answer:

1493.49

Step-by-step explanation:

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Please help me with 2b ASAP. <br> Really appreciate it!!

Bogdan [553]

$f(x)=\dfrac{x^2}{x^2+k^2}$

By definition of the derivative,

$f'(x)=\displaystyle\lim_{h\to0}\frac{\frac{(x+h)^2}{(x+h)^2+k^2}-\frac{x^2}{x^2+k^2}}h$

$f'(x)=\displaystyle\lim_{h\to0}\frac{(x+h)^2(x^2+k^2)-x^2((x+h)^2+k^2)}{h(x^2+k^2)((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{(x+h)^2-x^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2xh+h^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2x+h}{(x+h)^2+k^2}$

$f'(x)=\dfrac{2xk^2}{(x^2+k^2)^2}$

$\dfrac{k^2}{(x^2+k^2)^2}$ is positive for all values of $x$ and $k$ . As pointed out, $x\ge0$ , so $f'(x)\ge0$ for all $x\ge0$ . This means the proportion of occupied binding sites is an increasing function of the concentration of oxygen, meaning the presence of more oxygen is consistent with greater availability of binding sites. (The question says as much in the second sentence.)

7 0

3 years ago

Place 145 in the place value chart

Alexandra [31]

Answer: The 1 goes in the hundreds place and the 4 goes in the tens place and the 5 goes in the ones place

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please answer these 3 questions using this strategy.

34kurt

1. 4
2. 9
3. 21
are the answers

8 0

3 years ago

A rectangular prism has a length of 12in, a height of 5in, and a width of 8in. What is its volume, in cubic in?

cupoosta [38]

Answer: The volume is 480 in^3

Step-by-step explanation:

v= lwh

v= 12 * 5 * 8

v= 480 in^3

3 0

3 years ago

Read 2 more answers

Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$

Hence, the midpoint of the segment is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

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3 years ago