In ΔGHI, the measure of ∠I=90°, the measure of ∠H=52°, and HI = 63 feet. Find the length of IG to the nearest tenth of a foot.

podryga [215]

Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:

$\frac{d}{dx}(x^{2} - 1)^{3}$

So we apply chain rule:

= $3(x^{2} - 1)^{2} * 2x$

Set our first derivative to zero and solve for x:

3(x^2 - 1) * 2x = 0

So we can see that (by plugging in) 0, -1 and 1 makes our solution true

So our critical value is x = 0, x = -1, x = 1

5 0

3 years ago

If 4w - 1 = 11, then 4w = 12<br> what is the property?

KonstantinChe [14]

Answer:

Addition property of equality.

Hope this helps :)

7 0

3 years ago

A bag contains 5 red marbles , 2 blue marbles, 4 white marbles, and 1 yellow marble. if 2 marbles are chosen in a row , without

slega [8]

P of 1st white marble = 4/ 12

and P of 2nd white in a row = 3 /11

total = ( 4/12) * ( 3/11)
= 1/11

hope it helped

8 0

3 years ago

Dont really understand how to do this

bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

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

Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

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

The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$

The sum of the first ten terms is -85

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

As a check for $S_{10}$ , here are the first ten terms:

t1 = 5
t2 = 2
t3 = -1
t4 = -4
t5 = -7
t6 = -10
t7 = -13
t8 = -16
t9 = -19
t10 = -22

Then adding said terms gets us...

5 + 2 + (-1) + (-4) + (-7) + (-10) + (-13) + (-16) + (-19) + (-22) = -85

This confirms that $S_{10} = -85$ is correct.

6 0

1 year ago

Graph x^2/49+y+1^2/4=1

11Alexandr11 [23.1K]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

Perhaps you want a graph of ...

x^2/49 +(y +1)^2/4 = 1

This is an ellipse centered at (x, y) = (0, -1) with a major axis in the x-direction of 14, and a minor axis in the y-direction of 4.

3 0

3 years ago