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vesna_86 [32]
3 years ago
11

A roll of ribbon was 12 meters long. Diego cut 9 pieces of ribbon that were 0.4 meters each to tie some presents. He then used t

he remaining ribbon to make some wreaths. Each wreath required 0.6 meter.
How many meters of ribbon were available for making wreaths?

_____ meters

How many wreaths could Diego make with the available ribbon?

_______wreaths
Mathematics
2 answers:
Helen [10]3 years ago
7 0
Work and Reasoning:
We start off with 12m
Since there are 9 ribbons each 0.4 meters you have to do 9 • 0.4. Since that is 3.6m.
To find out how much ribbon you have left you must subtract the amount he used against the amount he had to begin with.
12 - 3.6 = 8.4
To find out how many wreaths he can make, you must divide the remaining amount of ribbon he has by the 0.6 meter.
8.4 divided by 0.6 is 14.

Answer:
He can make 14 wreaths with the remaining amount of ribbon.
dexar [7]3 years ago
4 0

Answer: 8.4; 14.

Step-by-step explanation:

To find how much ribbon we used to tie presents, we need to multiply 9 and 0.4.

9 x 0.4 = 3.6.

Diego used 3.6 meters of yard to tie some presents. We can subtract that from 12 to find how much ribbon we have left.

12 - 3.6 = 8.4.

We have 8.4 meters of ribbon to make wreaths.

Now we have to find how many wreaths we can make wiht 8.4 meters of ribbon. We know that each wreath needs 0.6, so we should divide 8.4 by 0.6 to find how many wreaths we can make.

8.4 ÷ 0.6 = 14.

Therefore, we can make 14 wreaths with the 8.4 meters of leftover ribbon.

(Part A's answer is 8.4, and Part B's answer is 14)

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7 0
3 years ago
Read 2 more answers
Car rentals involve a $130 flat fee and an additional cost of $31.67 a day. what is the maximum number of days you can rent a ca
Vladimir79 [104]
Represent the number of days by x. With this representation, the variable cost of the rental is 31.67x. The total cost is the sum of the fixed and variable costs. This value should not be more than $500. The equation below shows the relationship.
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Solving for x gives x ≤ 11.68
Thus, the maximum number of days to rent the car is only 11 days. 
6 0
3 years ago
Read 2 more answers
Please help explanation if possible
poizon [28]

Step-by-step explanation:

Hi there!

The given equation is:

y = -2x + 5………………(i)

Comparing the equation with y = mx+c, we get;

m1 = -2

Also another equation of the line which passes through point (-4,2), we get;

(y-2) = m2(X+4)............(ii) { using the formula (y-y1) = m2(x-x2)}

According to the question, they are perpendicular to eachother, So according to the condition of perpendicular lines;

m1*m2 = -1

-2*m2 = -1

or, m2 = 1/2.

Therefore, m2= 1/2.

Now, keeping the value of m2 in equation (ii).

(y-2) = 1/2(x+4)

y = (1/2)x + 4

Therefore, the required equation is: y = (1/2)x + 4.

<u>Hope</u><u> it</u><u> helps</u><u>!</u>

3 0
3 years ago
I got the first two can you help me with the last one PLZ
Nat2105 [25]

1. Geometric Sequence

2. a_n = a_{n-1} * 3

3. a_n = 6 * (3)^{n-1}

Step-by-step explanation:

Given sequence is:

6, 18, 54, 162,....

Here

a_1 =6\\a_2 = 18\\a_3 = 54

(a) Is this an arithmetic or geometric sequence?

We can see that the difference between the terms is not same so it cannot be an arithmetic sequence.

We have to check for common ratio (ratio between consecutive terms of a sequence) denoted by r

r = \frac{a_2}{a_1} = \frac{18}{6}= 3\\r = \frac{a_3}{a_2} = \frac{54}{18} = 3

As the common ratio is same, the given sequence is a geometric sequence.

(b) How can you find the next number in the sequence?

Recursive formulas are used to find the next number in sequence using previous term

Recursive formula for a geometric sequence is given by:

a_n = a_{n-1} * r

In case of given sequence,

a_n = a_{n-1} * 3

So to find the 5th term

a_5 = a_4*3\\a_5 = 162*3\\a_5 = 486

(c) Give the rule you would use to find the 20th week.

In order to find the pushups for 20th week, explicit formul for sequence will be used.

The general form of explicit formula is given by:

a_n = a_1 * r^{n-1}

Putting the values of a_1 and r

a_n = 6 * (3)^{n-1}

Hence,

1. Geometric Sequence

2. a_n = a_{n-1} * 3

3. a_n = 6 * (3)^{n-1}

Keywords: Geometric sequence, common ratio

Learn more about geometric sequence at:

  • brainly.com/question/10666510
  • brainly.com/question/10699220

#LearnwithBrainly

4 0
3 years ago
Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f
drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

y''+3y'=0

The characteristic equation is

r^2+3r=r(r+3)=0

with roots r=0 and r=-3, giving the two solutions C_1 and C_2e^{-3t}.

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

y''+3y'=2t^4

Assume the ansatz solution,

{y_p}=at^5+bt^4+ct^3+dt^2+et

\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e

\implies {y_p}''=20at^3+12bt^2+6ct+2d

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution C_1 anyway.)

Substitute these into the ODE:

(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4

15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4

\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}

y''+3y'=t^2e^{-3t}

e^{-3t} is already accounted for, so assume an ansatz of the form

y_p=(at^3+bt^2+ct)e^{-3t}

\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}

\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}

Substitute into the ODE:

(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}

9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2

-9at^2+(6a-6b)t+2b-3c=t^2

\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}

y''+3y'=\sin(3t)

Assume an ansatz solution

y_p=a\sin(3t)+b\cos(3t)

\implies {y_p}'=3a\cos(3t)-3b\sin(3t)

\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)

Substitute into the ODE:

(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)

(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)

\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}

So, the general solution of the original ODE is

y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}

3 0
3 years ago
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