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3 years ago

Find the radius, r, if the circumference is 10π. Formula C = 2πr

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

5 0

10π=2πr
Divide both sides by 2π
10π/2π=r
Simplify
5=r

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Hallar el valor de la incognita de la siguiente ecuacion 3(x-1)-4(5-x)=2(6+x)

Anton [14]

Answer;

x = 7

Step-by-step explanation:

3(x - 1) - 4(5 - x) = 2(6 + x)

3x - 3 -20 + 4x = 12 + 2x

7x - 23 = 12 + 2x

7x - 2x = 23 + 12

5x = 35

x = 35/5

x = 7

3 0

3 years ago

Can smbdy help me???

Sergio [31]

9514 1404 393

Answer:

a. One solution

Step-by-step explanation:

These are two linear equations with different x-coefficients. They have one solution.

<em>Additional comment</em>

If the x-coefficients are the same, the system may have 0 or infinite solutions, depending on the constants. It is not possible for a system of linear equations to have exactly two solutions.

3 0

3 years ago

Sharon kicks a ball from the ground into the air with an upward velocity of 64 feet per second. The function h = -16t2 + 64t mod

iren2701 [21]

When the ball reaches the ground is equal to one of the zeros, or x-intercepts. In order to find the zeros of a function, you need to factor the equation.
y=16x^2+64x
y=-16x(x-4)
The zeros are x=0 and x=4
Therefore the answer is D, after 4 seconds.
Hope this helps!

7 0

3 years ago

Read 2 more answers

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

HELP PICTURE IS SHOWN!!!

Liono4ka [1.6K]

Each outlier in a box plot is symbolised by a "dot". So there can be two possible answers since there are two different outliers in each end of the box plot (5 and 73): either C or D.

The line in the middle of the boxes is equal to the median. If you find the median, you can approximately determine where the middle-line is, and choose whether C or D is is best fit.

4 0

3 years ago