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sergiy2304 [10]
3 years ago
8

In circle O, radius OQ measures 9 inches and arc PQ measures 6π inches. Circle O is shown. Line segments P O and Q O are radii w

ith length of 9 inches. Angle P O Q is theta. What is the measure, in radians, of central angle POQ? StartFraction 2 pi Over 3 EndFraction radians StartFraction 3 pi Over 4 EndFraction centimeters StartFraction 4 pi Over 3 EndFraction radians StartFraction 3 pi Over 2 EndFraction radians
Mathematics
2 answers:
kupik [55]3 years ago
5 0

The answer is A. 2pi/3 radians.

nikklg [1K]3 years ago
4 0

Answer:

First Option

Explanation:

A)  2pi/3 radians

Hope this helps :)

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Find the sum of -44, -10, and 49​
Dmitriy789 [7]

Answer:

-5

Step-by-step explanation:

(-44) + (-10) + 49

=> 49 - 44 - 10

=> 49 - 54

=> -5

4 0
3 years ago
Read 2 more answers
Solve the following equation for h.<br> r+3Q/h=t
Arada [10]

Answer:

             \bold{h=\dfrac{3Q}{t-r}}

Step-by-step explanation:

\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}

Or, if you mean (r+3Q)/h=t:

\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}

4 0
3 years ago
What is the average rate of change of f(x) = -x2 + 3x + 6 over the interval –3
Rufina [12.5K]

Answer:

\frac{\Delta y}{\Delta x}  =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3

Step-by-step explanation:

To find the average rate of change of a function over a given interval, basically you need to find the slope. The mathematical definition of the slope is very similar to the one we use every day. In mathematics, the slope is the relationship between the vertical and horizontal changes between two points on a surface or a line. In this sense, the slope can be found using the following expression:

Average\hspace{3}rate\hspace{3}of\hspace{3}change=Slope=\frac{y_2-y_1}{x_2-x_1}  =\frac{f(x_2)-f(x_1)}{x_2-x_1}

So, the average rate of change of:

f(x)=-x^2+3x+6

Over the interval -3

Is:

f(x_2)=f(3)=-(3)^2+3(3)+6=-9+9+6=6\\\\f(x_1)=f(-3)=-(-3)^2+3(-3)+6=-9-9+6=-12

\frac{\Delta y}{\Delta x}  =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3

Therefore, the average rate of change of this function over that interval is 3.

3 0
3 years ago
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
Will give brainliest + 5 star rating!
koban [17]

Answer is choice C) 4x^2-3x+9

You list off the terms in a way where the largest exponent term comes first, then the second largest exponent is next, and so forth. Note how the last term 9 is the same as 9x^0. The term 3x is equivalent to 3x^1. So what we really have is 4x^2-3x^1+9x^0. The exponents count down: 2, 1, 0

3 0
4 years ago
Read 2 more answers
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