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DIA [1.3K]
3 years ago
9

Help with 10-14 questions plz

Mathematics
1 answer:
lesya [120]3 years ago
3 0
#10) 63,945 cubic inches.
#11) PRISM A = 120 cubic inches. PRISM B = 960 cubic inches. Prism B is 8 times greater in size.
#12) 58.4375 cubic centimeters.
#13) 58.5 inches.
#14) 3,860 grams.
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How many mL are there in 3,700 microliters?
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Answer:

3.7 mL

Step-by-step explanation:

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The universal set in this diagram is the set of integers from 1 to 15. Place the integers in the correct place in the Venn Diagr
Dmitry_Shevchenko [17]

Answer:

see image

Step-by-step explanation:


7 0
3 years ago
a video game has 35 levels in it if you beat 5 of the levels what is the ratio of levels left to the levels that have been beate
marta [7]

We have 35 - 5 = 30 levels left to beat.

The question is asking us for the levels left to beat, to the ones that already have been beaten:

30 : 5

Simplified;

6 : 1

8 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}.

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3 years ago
Shawn is collecting money from students at his school for charity the table gives the ratio of the number of students who have d
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