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2 years ago

I'm new to this math and it's very confusing it's asking that Fill out the Venn diagram using the given information.

Mathematics

1 answer:

GREYUIT [131]2 years ago

5 0

There are the answers filled in :)

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Three improper fractions for 4 1/2

Brut [27]

9/2, 18/4, 27/6 you just make a equivalent fraction then multiply the whole number by the denominator and add the numerator

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2 years ago

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Identify tan P as a fraction and as a decimal rounded to the nearest hundredth. HELP PLEASE!!

irakobra [83]

$tangent=\dfrac{opposite}{adjacent}$

We have:

$opposite=8.5\\adjacent=16$

Substitute:

$\large\boxed{\tan P=\dfrac{8.5}{16}\approx0.53}$

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3 years ago

I need help with b and c please and thank you!!

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2 years ago

Sorry to waste time, but could you help? 5/3(4/3a+9b+2/3a) simplified, take your time! I’m not the best at math so :p

mixas84 [53]

Answer:

sorry I'm not that good at math either it's complicated

Step-by-step explanation:

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⋅

5 0

3 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago