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2 years ago

Eighth grade > V.4 Area and circumference of circles CHV

Mathematics

1 answer:

kati45 [8]2 years ago

8 0

Answer: $D=13.8$

Step-by-step explanation:

To fing the diameter of a circle when given the circumfrence we use the formula of

$D=\frac{c}{\pi }$

$D= \frac{43.332}{3.14}$

$D=13.8$

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The graph of r = 1/ 1-5 sin theta is symmetric about the

Gnoma [55]

Answer:

A. y-axis only

Step-by-step explanation:

We have the graph of as given below.

It can be seen that passes through the points (0,0), (4,0) and (6,0).

Also, the graph is plotted in the 1st and 2nd quadrant symmetrically along the y-axis.

Hence, the axis of symmetry of is the y-axis only.

7 0

3 years ago

Read 2 more answers

Greatest common factor of 66 and 32

scoundrel [369]

Hey there,

The Greatest Common Factor or the GCF of 66 and 32 would be 2.

Both of the numbers go into 2 and that is the highest common factor the two share.

Hope I helped,

Amna

8 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Solve the inequality 5x<2x-3

mr Goodwill [35]

Answer:

To solve this inequality you need to get x by itself

5x<2x-3 (your first step would be to subtract 2x from each side)

3x<3 ( you then divide 3 to each side to get x by itself)

x<1

Hope this helps ;)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Is -134 greater than or less than 1136

masya89 [10]

Answer:

less than.

Step-by-step explanation:

.............

7 0

3 years ago