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3 years ago

I need an example of a proportional relationship

Mathematics

2 answers:

nikitadnepr [17]3 years ago

3 0

Answer:

if you want an equation... an example could be

y=3x+5

Step-by-step explanation:

Colt1911 [192]3 years ago

3 0

Answer:

Proportions are the same ratios written in different forms. A proportional relationship is states that they are the same. For example, 1/2 and 6/12 have a proportional relationship, which means they are the same.

pls give brainliest

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80% of Vanessa's final grade is based on the average score of her quizzes, which is 73.

Misha Larkins [42]

80% of quizzes + 20% of final = final grade

80% * 73 + 20% * x = 70

0.8 * 73 + 0.2x = 70

58.4 + 0.2x = 70

0.2x = 11.6

x = 58

8 0

4 years ago

Write an equivalent ratio for 3/30

nalin [4]

Answer:

One example of an equivalent ratio to 3/30 is 1:10.

6 0

4 years ago

Which number represents the mode of the data? a) 100 b) 150 c) 200

luda_lava [24]

Answer:

200

Step-by-step explanation:

Mode is the highest frequency of <em>1</em> number. In this case, 200 shows up the most. Therefore, your answer is C.

5 0

4 years ago

Read 2 more answers

Who invented the square root of negative one?

telo118 [61]

I believe it would be jack Dorsey and jim Mckelvey since they were the ones who made the perfect square

6 0

4 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago

I need an example of a proportional relationship​

I need an example of a proportional relationship