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3 years ago

Adding and subtracting fractions with unlike denominators

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

If the Problem wa 2/10+2/100 you would do this 10/10 +2/10= 20/100 and add that new number

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Use Modus Ponens to deduce the conclusion from each of the following pairs of premises:

Nuetrik [128]

The conclusion is a=c statement

A modus ponens argument has the same structure as a syllogism, with two premises and a conclusion:

If P, then Q.

P occurs.

Consequently, Q occurs.

The Modus Ponens rule of inference or rule of logic requires a single premise and its logical consequences. A conditional statement states that if event 1 occurs, event 2 will also occur and that event 2 will be inferred as the outcome if event 1 occurs. For instance, if A implies B and A is assumed to be true, then it follows that B must also be true according to the Modus Ponens rule.

Hence, By modus Ponens

a=b∧b=c ⇒ a=c

So, given a=b∧b=c Then

⇒a=c.

Hence the conclusion is a=c statement .

Learn more about Modus ponens here-

brainly.com/question/15268011

#SPJ10

5 0

2 years ago

Ian invests $13,670 in a savings account at his local bank which gives 1.9% simple annual interest. He also invests $6,040 in an

IceJOKER [234]

The formula says
I=prt
I Interest earned
P principle
R interest rate
T time

The interest earned of local account
13,670×0.019×9
=2,337.57

The interest earned of online account
6,040×0.045×9
=2,446.2

The difference
2,446.2−2,337.57
=109

So the answer is d

7 0

3 years ago

Read 2 more answers

Correct answer will get Brainliest.

IRINA_888 [86]

Answer:

Step-by-step explanation:

They can't form a triangle because the sum of 2 sidelengths is ALWAYS greater than the 3rd side in a triangle. Because 5+7=12<14, this means that it can't form a triangle.

3 0

3 years ago

Express the ratio of 3 ft to 18 in lowest terms

seraphim [82]

$1\ ft=12\ in\to3\ ft=3(12\ in)=36\ in\\\\3\ ft:18\ in=36\ in:18\ in=36:18=\dfrac{36}{18}=\dfrac{2}{1}=2:1\\\\Answer:\ \boxed{3\ ft:18\ in=2:1}$

4 0

4 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago