chong earns an annual salary of 29,500. how much is his biweekly paycheck? $1,134.62 $1,135.62 $1,234.62 none of the above

1 answer:

6 0

To figure this out, divide the annual salary by the months in a year.
29,500/12=2458.33
Now divide that by half of the amount of weeks in month
2458.33/1.5=1638.88
So the answer is none of the above.

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Please help!!

sveta [45]

For this case we have the following equation:
$d = \sqrt{\frac{3h}{2}}$
Where,
d: the distance they can see in thousands
h: their eye-level height in feet
For Kaylib:
$d = \sqrt{\frac{3(48)}{2}}$
$d=\sqrt{3(24)}$
$d=\sqrt{72}$
$d=6\sqrt{2}$
For Addison:
$d = \sqrt{\frac{3(85\frac{1}{3})}{2}}$
$d = \sqrt{\frac{256}{2}}$
$d=\sqrt{128}$
$d=8\sqrt{2}$
Subtracting both distances we have:
$8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2}$
Answer:
$2\sqrt{2}$
B. 2√2 mi

6 0

3 years ago

Read 2 more answers

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.