Ask question

Ask question

All categories

3 years ago

7

Solve the following dividend problem.alameda tent company has 75,000 shares of stock outstanding. what total dividend declaratio

n would be necessary for the dividend per share to be $2.00.

2 answers:

butalik [34]3 years ago

5 0

75,000×2=150,000
...............

Otrada [13]3 years ago

4 0

Answer:

150.000

Step-by-step explanation:

You might be interested in

What is 25% of 60 dhhsbebebebrbrbrnr

weqwewe [10]

Answer:

15

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Can someone pls tell me how to graph to find the solution?

Bas_tet [7]

Answer:

Rearrange the equation so "y" is on the left and everything else on the right.

Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)

Shade above the line for a "greater than" (y> or y≥) or below the line for a "less than" (y< or y≤).

Step-by-step explanation:

sorry if wrong

3 0

3 years ago

Need help asap need it thenext few minutes

pantera1 [17]

Answer:

Correct Option is d $(x+3)^2 = 11$

Step-by-step explanation:

For completing the square our equation should be in the form of $a^2 +2ab + b^2 = (a+b)^2$

In the given equation we have:

$x^2 +6x -2\\x^2 + 2(x) (?) +(?)^2 = 2\\for\,\, making\,\, 6x\,\, 2*x*3=6x \\so, \,\,we\,\, can\,\, add\,\, and\,\, subtract\,\, (3)^2 \,\,on\,\, both\,\, sides\\x^2 + 2(x) (3) +(3)^2 -(3)^2= 2\\(x+3)^2 -9 =2\\(x+3)^2 =2+9\\(x+3)^2 = 11$

Correct Option is d $(x+3)^2 = 11$

6 0

3 years ago

For what values of θ is sinθ < cosθ when 0 ≤ θ < π ?

zysi [14]

<h3>Answer: $0 \le \theta < \frac{\pi}{4}$ </h3>

=========================================================

How to get this answer:

Use the unit circle to note that $\sin\theta = \cos\theta = \frac{\sqrt{2}}{2}$ when $\theta = \frac{\pi}{4}$ (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$

-------------------------------

Here's a more detailed picture of whats going on.

$\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\$

Intersect the intervals $0 \le \theta < \pi$ and $\theta < \frac{\pi}{4}$ and you'll end up with the final answer $0 \le \theta < \frac{\pi}{4}$

4 0

3 years ago

Please solve number 16 for me.

vodka [1.7K]

Answer:

post pictures also for your answer

4 0

3 years ago