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BlackZzzverrR [31]

2 years ago

14

If the price of a certain stock is $66.50 per share and its earnings are $4.75 per share, what is the ratio of the stock’s price

to its earnings in lowest terms?

1 answer:

Blababa [14]2 years ago

5 0

Answer:

I may be wrong but I think it's 14

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Please tell me where to graph this!! (include if i open circle or not please) (NO FILE ANSWER!!!)

3241004551 [841]

Answer:

1.Make the circle open

2.put the dot on 5

3. change direction

Step-by-step explanation:

A solid dot on a number line graph indicates that the given number should be included as a possible solution, whereas an open dot indicates that the given number cannot be a solution. For example, if you graph x > 7, you place an open dot at 7.

Have a nice day! :-)

7 0

3 years ago

Read 2 more answers

My professor showed us in class today how to use ode45 to solve a differential equation numerically. i would like to use it on m

Brilliant_brown [7]

Yes, ode45 can be used for higher-order differential equations. You need to convert the higher order equation to a system of first-order equations, then use ode45 on that system.

For example, if you have

... u'' + a·u' + b·u = f

you can define u1 = u, u2 = u' and now you have the system

... (u2)' + a·u2 + b·u1 = f

... (u1)' = u2

Rearranging, this is

... (u1)' = u2

... (u2)' = f - a·u2 - b·u1

ode45 is used to solve each of these. Now, you have a vector (u1, u2) instead of a scalar variable (u). A web search regarding using ode45 on higher-order differential equations can provide additional illumination, including specific examples.

7 0

3 years ago

For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

natulia [17]

<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let $e$ the cost of the exam hours

Let $w$ be the cost of the workshop hours

Let $l$ be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

$e=2w$ equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

$w=2l$ equation (2)

Finally, we also know that 3hr exams 24hr workshops and 12hr lectures cost $528, so:

$3e+24w+12l=528$ equation (1)

Now, lets find the value of $l$ :

Step 1. Solve for $l$ in equation (3)

$3e+24w+12l=528$

$12l=528-3e-24w$ equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

$12l=528-3e-24w$

$12l=528-3(2w)-24w$

$12l=528-6w-24w$

$12l=528-30w$ equation (5)

Step 3. Replace equation (2) in equation (5) and solve for $l$

$12l=528-30w$

$12l=528-30(2l)$

$12l=528-60l$

$72l=528$

$l=\frac{528}{72}$

$l=\frac{22}{3}$

$l=7.33$

Cost of lectures = $7.33 per hour

3 0

3 years ago

Read 2 more answers

If the area of a circle is 58 square feet, find the circumference. A. 42.5 ft. B. 4.25 ft. C. 22.35 ft. D. 26.99 ft.

PSYCHO15rus [73]

Answer: OPTION D.

Step-by-step explanation:

The formula for calculate the circumference of a circle is:

$C=2r\pi$

Where <em>r</em> is the radius.

The formula for calculate the area of a circle is:

$A=r^2\pi$

Where <em>r</em> is the radius.

Solve for <em>r</em> from $A=r^2\pi$ to calculate it:

$r=\sqrt{\frac{A}{\pi}}\\r=\sqrt{\frac{58ft^2}{\pi}}\\r=4.296ft$

Subsitute the radius into $C=2r\pi$ . Then:

$C=(2)(4.296ft)\pi=26.99ft$

3 0

3 years ago

Read 2 more answers

(x+5)(√ 1-x)<br> what is the product of this composite function?

Radda [10]

Answer:

$x\sqrt{1-x}+5\sqrt{1-x}$

Step-by-step explanation:

$(x+5)(\sqrt{1-x} )\\\\=\left(\sqrt{1-x}\right)\left(x+5\right)\\\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\\\a=\left(\sqrt{1-x}\right),\:b=x,\:c=5\\\\=\left(\sqrt{1-x}\right)x+\left(\sqrt{1-x}\right)\times\:5\\\\=x\left(\sqrt{1-x}\right)+5\left(\sqrt{1-x}\right)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\\\=x\sqrt{1-x}+5\sqrt{1-x}$

6 0

3 years ago