Im confused please help

zlopas [31]

The first, fourth and fifth ones are linear.

The second, third and sixth are all nonlinear.

You can tell this by the fact that their lead exponents are not 1.

3 0

3 years ago

What are the leading coefficient and degree of the polynomial? 3u+12u'-7u* +1

xxTIMURxx [149]

Answer:

See explanation

Step-by-step explanation:

Assuming the given polynomial is

$3u + 12 {u}^{2} - 7 {u}^{3} + 1$

We then rewrite in decreasing powers of u, to get:

$- 7 {u}^{3} + 12 {u}^{2} + 3u + 1$

The leading term is

$- 7 {u}^{3}$

The leading coefficient is the coefficient of the leading term.

The leading coefficient is -7.

6 0

3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

2 years ago

An artist wants to frame a square painting with an area of 400 square inches if x is the length of the one side of the painting

tatuchka [14]

There is only one solution to the equation. A quadratic equation has 2 solutions a positive and a negative, but since we are looking for length, the answer can only be postive. Area= side^2

400= x^2

sqrt(400)=x

20 inches = x

The length of one side is 20 inches

check: 20*20 = 400, which is the area of the square painting.

8 0

3 years ago

How am I supposed to figure this out if csc Theta(0) has a radical number on the top?

Harman [31]

$\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)=\cfrac{\sqrt{7}}{2}\cfrac{\leftarrow hypotenuse=c}{\leftarrow opposite=b}\\\\
-----------------------------\\\\
\textit{so hmm using the pythagorean theorem, we can say that}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
b=opposite\\
a=adjacent
\end{cases}\\\\
\textit{we'll assume is the + of the root, since you're not}\\
\textit{given further info on it, that means the I quadrant}\\\\
$

$\bf -----------------------------\\\\
cos(\theta)=\cfrac{a}{c}\qquad sec(\theta)=\cfrac{c}{a}\qquad tan(\theta)=\cfrac{b}{a}\qquad cot(\theta)\cfrac{a}{b}
$

so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant

keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant

but anyhow, use the I quadrant

7 0

2 years ago