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2 years ago

Is 56/8 an interger Please help....

Mathematics

1 answer:

kkurt [141]2 years ago

5 0

Fractions and decimals cannot be integers. whole, natural numbers are integers (ex. 5,6,7).

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Please see the image and help with this homework problem. thank you!

dimulka [17.4K]

we have the function

$E=\sin \frac{\pi}{14}t$

Part a

For t=7

substitute in the given function

$\begin{gathered} E=\sin \frac{\pi}{14}7 \\ E=1 \end{gathered}$

For t=14

$\begin{gathered} E=\sin \frac{\pi}{14}14 \\ E=0 \end{gathered}$

For t=21

$\begin{gathered} E=\sin \frac{\pi}{14}21 \\ E=-1 \end{gathered}$

For t=28

$\begin{gathered} E=\sin \frac{\pi}{14}28 \\ E=0 \end{gathered}$

For t=35

$\begin{gathered} E=\sin \frac{\pi}{14}35 \\ E=1 \end{gathered}$

Observation: The values of E varies from -1 to 1, including the zero

Part B

Remember that

The Period goes from one peak to the next

Period=2pi/B

B=pi/14

Period=(2pi)/(pi/14)=2pi*14/pi=28

<h2>the period is 28 days</h2>

6 0

1 year ago

Find the greatest common factor of 15x^2y^3 and -18x^3yz <br> The Answer has to be like _x_y_

Daniel [21]

Answer:

The greatest common factor of this would be 3x^2y

Step-by-step explanation:

In order to find this, first find the greatest common factor of the coefficients. Since 3 goes in evenly to both 15 and -18, then we know that it is a common factor.

From there we need to find the number of x's. Since the first term only has 2 x's and the second has 3, we take the lowest number. (x^2)

And since the first term has 3 y's and the second has just 1, we take the lowest number (y).

8 0

3 years ago

Lines e and f are parallel. The mAngle9 = 80° and mAngle5 = 55°. Parallel lines e and f are cut by transversal c and d. All angl

brilliants [131]

Answer:

$(A)m\angle 2=125^\circ\\(C)m\angle 8=55^\circ\\(E)m\angle 14=100^\circ$

Step-by-step explanation:

The diagram of the problem is drawn and attached.

Given that:

$m\angle 9 =80^\circ\\m\angle 5 =55^\circ$

$m\angle 5+ m\angle 6=180^\circ\\55^\circ+ m\angle 6=180^\circ\\m\angle 6=180^\circ-55^\circ\\m\angle 6=125^\circ\\$Now:\\m\angle 6 = m\angle 2 $(Corresponging angles)\\Therefore m\angle 2=125^\circ\\$

$m\angle 5= m\angle 8=55^\circ $(Vertically Opposite Angles)$

Also

$m\angle 9+ m\angle 10=180^\circ\\80^\circ+ m\angle 10=180^\circ\\m\angle 10=180^\circ-80^\circ\\m\angle 10=100^\circ\\$Now:\\m\angle 10 = m\angle 14 $(Corresponging angles)\\Therefore m\angle 14=100^\circ\\$