The sum of the polynomials 6x3 8x2 – 2x 4 and 10x3 x2 11x 9 is

GarryVolchara [31]

Answer:

Try the gauth math app it's very useful to use

5 0

2 years ago

The minute hand of a clock is exactly 6in in length, and the tip of the minute hand has traveled ten inches since noon. What tim

frutty [35]

$\bf \textit{arc's length}\\\\
s=\cfrac{\theta \pi r}{180}\qquad 
\begin{cases}
r=radius\\
\theta =angle\ in\\
\qquad degrees\\
------\\
r=6\\
s=10
\end{cases}\implies 10=\cfrac{\theta \pi 6}{180}\implies \cfrac{180\cdot 10}{6\pi }=\theta 
\\\\\\
\cfrac{300}{\pi }=\theta \implies 95.49^o\approx \theta$

now, the circle of the clock has 360°, if we divide it by 60(minutes), we get 360/60, just 6° for each minute.

now, if there are 6° in 1 minute, how many minutes in 95.49°?

well, just 95.49/6 or about 15.92 minutes, I take it you can round it up to 16 minutes.

so 16 minutes since noon, so is about 12:16, about time get the silverware for lunch.

3 0

3 years ago

PLEASE HELP IM STUCK

uysha [10]

Y axis and quadrant IV

6 0

2 years ago

Which of the following represents The linear equation 3x+y=2 in slope-intercept form

ollegr [7]

C, see explanation in the picture attached below.

4 0

2 years ago

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = - $\sqrt{3}$

cos = $\frac{-\sqrt{3}}{3}$

Hope this helps!! :)

3 0

2 years ago

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