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olya-2409 [2.1K]
3 years ago
8

-9 is greater than: A. -10 B. -7 c. -6 D. -9 E. O

Mathematics
2 answers:
garri49 [273]3 years ago
7 0

Answer: A. -10

Step-by-step explanation:

GaryK [48]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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A baseball diamond is a square whose sides are 9090 feet long. Suppose a player running from second base to third base has a spe
Arada [10]
<h2>Answer:</h2>

<em><u>Rate of change of distance at that instant = 0.066 ft/sec </u></em>

<h2>Step-by-step explanation:</h2>

In the question,

The baseball field is in the square diamond shape.

Side length of square = 9090 ft.

Now,

The player runs from the 2nd base to the 3rd base.

Speed of running of player, v = 30 ft/ sec.

Distance of the player from third base = 20 ft.

Now,

Let us say the distance of the player from the Home base is = l

So,

In the triangle using the Pythagoras theorem, we get,

l^{2}=x^{2}+s^{2}

where, 'x' is the distance of the player from the third base and 's' is the side of the square field base.

So,

l^{2}=x^{2}+s^{2}\\On\,differentiating\,we\,get,\\2l\frac{dl}{dt}=2x\frac{dx}{dt}\\l\frac{dl}{dt}=x\frac{dx}{dt}\\

Now,

Also, at the moment when, x = 20,

Length from the Home base is,

l^{2}=20^{2}+9090^{2}\\l=9090.022\,ft.

Now,

On putting we get,

l\frac{dl}{dt}=x\frac{dx}{dt}\\(9090.022)\frac{dl}{dt}=20(30)\\\frac{dl}{dt}=0.066\,ft./sec.

<em><u>Therefore, the rate of change of distance from the home plate is 0.066 ft/s</u></em>

6 0
4 years ago
COUGH COUGH CORONA I HAVE CORONA HERE SOME POINTS
Helga [31]
Thanks for the points lol
6 0
3 years ago
Help I don’t get it at all
Naily [24]
The answer to your question is n is equal to 70
5 0
3 years ago
Read 2 more answers
What is 2y + x (3y) = - 15
enot [183]

Answer:

x=\frac{-2y-15}{3y}

y=\frac{-15}{3x+2}

4 0
4 years ago
What is the confidence interval estimate of the population mean percentage change in the price per share of stock during the fir
I am Lyosha [343]

Answer:

(-0.1059 ; - 0.0337)

Step-by-step explanation:

The data table is attached in the picture below:

These is a matched pair design ; which requires taking the difference of the two values for each sample :

The mean and standard deviation of the difference will be used to construct the confidence interval :

The mean of difference, dbar = Σx/n = - 0.0698

The standard deviation of difference, Sd ;

Sd = [√Σ(d - dbar)²/(n-1)] = 0.1054

n = sample size = 25

The confidence interval :

dbar ± [TCritical * Sd/√n]

Tcritical at 90% ; df = n -1 = 25 -1

Tcritical(90% , 24) = 1.1711

C.I = - 0.0698 ± (1.711 * 0.1054/√25)

C.I = - 0.0698 ± 0.0361

C.I = (-0.1059 ; - 0.0337)

5 0
3 years ago
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