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3 years ago

What is 16/2 as a mixed number?

Mathematics

2 answers:

Rasek [7]3 years ago

7 0

Answer: 8

Step-by-step explanation:

Colt1911 [192]3 years ago

5 0

Answer: 8

Step-by-step explanation: If you do the same process you are still going to get 8 because there is no remainder like in some other fractions like 4/3

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The following set of ordered pairs represent a function:

MAXImum [283]

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3 years ago

(Please help, due by midnight PST in 2hours) Find x and y.

bezimeni [28]

X= 20 y= 100
I hope it helped. Give Brainly if it help

5 0

3 years ago

A taxi charges a flat rate of $1.75, plus an additional $0.65 per mile. If Erica has at most $10.20 to spend on a

JulijaS [17]

Answer:13 miles

Step-by-step explanation:10.20-1.75=8.45÷0.65=13

8 0

3 years ago

An arc with length 5 pi over 4 inches is formed by a 15 degree central angle what is the radius of the circle

Natalka [10]

Answer:

Radius of the circle is 15 inches.

Step-by-step explanation:

Relation between length of arc, radius and the angle subtended by the arc on center is:

$\theta = \dfrac{l}{r} ..... (1)$

where $\theta$ is the central angle in radians subtended by arc

$l$ is the length of arc

$r$ is the radius of arc

We are Given the following details:

$l = \dfrac{5\pi}{4}\ inch$

$\theta = 15^\circ$

We know that $\pi \ radians = 180 ^\circ$

Converting $\theta$ to radians:

$\theta =\dfrac{\pi}{180} \times 15$ radians

Putting the values of $\theta$ and $l$ to find the value of $r$

$\dfrac{\pi}{180} \times 15 = \dfrac{5\pi}{4r}\\\Rightarrow r = \dfrac{45}{3} \\\Rightarrow r = 15 \ inches$

Hence, Radius of the circle is 15 inches.

6 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago