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3 years ago

Rebecca has 6 yards of ribbon. It takes 3/8 yard to wrap one package. How many packages can rebecca wrap

Mathematics

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

Step-by-step explanation:

6 x 8 = 48

48 / 3 = 16

Rebecca can only wrap 16 packages with 6 yards of ribbon.

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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

to avoid a large shallow reef a ship set a course from point a and traveled 24 miles east to point b . the ship them turn and tr

Sever21 [200]

First, we are going to find the distance traveled by the ship adding the tow distances:
Distance traveled= 24 mi +33 mi=57 mi

Next, we are going to use the Pythagorean theorem to find the distance from $a$ to $c$ :
$d^2=24^2+33^2$
$d^2=576+1089$
d^2=1665
$d= \sqrt{1665}$
$d=40.8$ mi

Finally, we are going to subtract the two distances:
57 mi -40.8 mi= 16.2 mi

We can conclude that <span>if the ship could have traveled in a straight lime from point a to point c, it could have saved 16.2 miles.</span>

7 0

3 years ago

Basir buys 4 small drinks for $6.write and equation represent the cost c, for d small drinks?

maria [59]

Answer:

c = 1.5 d

Step-by-step explanation:

cost of 4 small drinks = $6

cost of 1 small drink = $ $\frac{6}{4}$ = $1.5

cost of d small drinks = 1.5 d = c

∴ c = 1.5 d

3 0

3 years ago

An image point has coordinates B'(7, -4). Find the coordinates of its pre-image if the translation used was (x + 4, y + 5).

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

B'(7, -4)

Let the pre-image coordinates be (x,y)

Consider x-coordinate,

x + 4 = 7

x = 3

Consider y-coordinates,

y + 5 = -4

y = -9

pre-image coordinates is (3, -9)

8 0

3 years ago

Read 2 more answers

A roll of paper towels is wound around a hollow cardboard tube. the

blsea [12.9K]

The radius of the cardboard tube of 85th loop of paper is 4.92cm

The arithmetic sequence is the sequence where every term is increased or decreased by a fixed number from the previous number.

Here the outer radius of the tube is 2.4 cm

the thickness of the paper is 0.3mm= 0.03cm

i.e. in every loop the increase in the radius of the loop is 0.03cm

then the radius in every sequence will be 2.40, 2.43, 2.46, 2.49, 2.52, .....

so here it is clear that it is an arithmetic sequence with a common difference of 0.03.

nth term of the sequence, aₙ = a₁ + (n - 1)d where a₁ is the first term, n is the index of the loop, and d is a common difference.

here a₁ =2.40

d=0.03

n=85

the radius of tube of the 85th loop will be= r= 2.40+(85-1)0.03= 2.40+ 2.52= 4.92

Therefore The radius of the cardboard tube of the 85th loop of paper is 4.92cm

Learn more about the arithmetic sequence

here: brainly.com/question/6561461

#SPJ10

6 0

2 years ago

Rebecca has 6 yards of ribbon. It takes 3/8 yard to wrap one package. How many packages can rebecca wrap​

Rebecca has 6 yards of ribbon. It takes 3/8 yard to wrap one package. How many packages can rebecca wrap