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2 years ago

The radius of a circle is 9in. Find it’s circumference in terms of

Mathematics

1 answer:

Lady bird [3.3K]2 years ago

3 0

${ \bf{ \underbrace{Given :}}}$

Radius of the circle " $r$ " = 9 in.

${ \bf{ \underbrace{To\:find:}}}$

The circumference of the circle.

${ \bf{ \underbrace{Solution :}}}$

$\sf\orange{The\:circumference \:of\:the\:circle\:is\:56.52\:in.}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

We know that,

$\sf\purple{Circumference\:of\:a\:circle \:=\:2πr }$

$= 2 \times 3.14 \times 9 \: in \\ \\ = 56.52 \: in$

Therefore, the circumference of the circle is 56.52 in.

$\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♡}}}}}$

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Which is defined using the undefined terms point and lines

Luba_88 [7]

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3 years ago

Read 2 more answers

4×-6+x+3 what is the value of this expression when x =8

soldi70 [24.7K]

Answer:

$- 13$

Step-by-step explanation:

$4 \times - 6 + x + 3$

GIVEN THAT,

$x = 8$

LET'S SOLVE NOW

$4 \times - 6 + 8 + 3 \\ - 24 + 8 + 3 \\ - 24 + 11 \\ = - 13$

Hope this helps

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2 years ago

In a baseball tournament, teams get 5 points for a win, 3 points for a tie and 1 point for a loss Nathan’s team has 29 points ho

puteri [66]

Answer:

37 different combinations

Step-by-step explanation:

First of all, we will count the possible combinations that add up to 29.

-Losses only;

One possibility: 29 losses

-Ties & losses;

Nine possibilities: 1 tie and 26 losses; 2 ties and 23 losses; 3 ties and 20 losses;4 ties and 17 losses; 5 ties and 14 losses; 6 ties and 11 losses; 7 ties and 8 losses; 8 ties and 5 losses; 9 ties and 2 losses

-Wins & losses

Five possibilities: 1 win and 24 losses; 2 wins and 19 losses; 3 wins and 14 losses; 4 wins and 9 losses; 5 wins and 4 losses

Now, what we want to find from the question is number of possibilities for wins, ties, and losses all together. So, we will count up the ties and losses in the remainder for each case of a given number of wins.

Thus;

For 0 Wins: 0 ties and 29 losses; 1 tie and 26 losses; 2 ties and 23 losses; 3 ties and 20 losses;4 ties and 17 losses; 5 ties and 14 losses; 6 ties and 11 losses; 7 ties and 8 losses; 8 ties and 5 losses; 9 ties and 2 losses.

Which sums up to 10 possibilities

For 1 win; 0 ties and 24 losses; 1 tie and 21 losses........8 ties and 0 losses.

Which sums up to 9 possibilities

For 2 wins; 0 ties and 19 losses; 1 tie & 16 losses............ 6 ties and 1 loss.

Which sums up to 7 possibilities

For 3 wins; 0 ties and 14 losses; 1 tie and 11 losses ....... 4 ties and 2 losses.

Which sums up to 5 possibilities

For 4 wins; 0 ties & 9 losses; 1 tie and 6 losses....... 3 ties and 0 losses

Which sums up to 4 possibilities

For 5 wins; 0 ties and 4 losses; 1 tie and 1 loss

Which sums up to 2 possibilities

Thus;

Total number of possibilities of combinations of wins, ties and losses = 10 + 9 + 7 + 5 + 4 + 2 = 37

8 0

3 years ago

If u(x) =-2x^2 and v(x)=1 over x what is the range of (uov) (x)

Radda [10]

u(x) = -2x², v(x)= 1/x

(u o v)(x) = u(v(x)) = -2(1/x)²= -2/x²

We can see that domain for x is going to bee all real numbers except 0. From the equation above we can see that graph of the function (u o v)(x) = -2/x² has horizontal asymptote y=0, because degree of numerator is less than degree of denominator ( (u o v)(x) = -2x⁰/x² ).

x² is always going to be positive, so range is going to be all negative numbers.

y<0, or y∈(0,-∞)

6 0

3 years ago

Which of the following would increase the width of a confidence interval for a population mean? Choose the correct answer below

Lelechka [254]

Answer:

A. Increase the level of confidence

Step-by-step explanation:

The margin of error is given by:

The margin of error is:

$M = \frac{Ts}{\sqrt{n}}$

In which T is related to the level of confidence(the higher the level of confidence, the higher T is), s is the standard deviation of the sample and n is the size of the sample.

Increase the width:

That is, increasing the margin of error, as the width is twice the margin of error, the possible options are:

Increase T -> increase confidence level.

Increase s -> Increase the standard deviation of the sample.

Decrease n -> Decrease the sample size.

Thus, the correct answer is given by option A.

7 0

2 years ago