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Fynjy0 [20]
3 years ago
5

D + m + 20=? .........................

Mathematics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

this cant be solved without more info

Step-by-step explanation:

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Doreen lives 3/5 mile from the library. Sheila lives 1/3 as far away from the library as Doreen. How far does Sheila live from t
tankabanditka [31]

Answer:

She lives 1/5 miles away from the library

Step-by-step explanation:

Here, we want to get the distance that Sheila

lives from the library

From the question, we are told she lives 1/3 as far as Doreen

So what we have to do in this case is to simply multiply the distances

we have this as:

1/3 * 3/5 = 1/5

6 0
4 years ago
How to tell if an equation is independent dependent or inconsistent?
Keith_Richards [23]
The example for an inconsistent system of equations:
x + 2 y = 10 and x + 2 y = 20. If we try to solve it algebraically:
x = 10 - 2 y; 10 - 2 y + 2 y = 20; 10 = 20 ( not true ). We will with something that is not true. Graph of those equations shows 2 parallel lines. Therefore the system is inconsistent.
The system is dependent when we have equivalent equations. For example:
2 x + y = 5 and 4 x + 2 y = 10. If we try to solve this system, we will end up with: 0 = 0 ( true ). The graph of this system shows the same line. This system is dependent.
Finally, if we have just one solution for the system of equations then the system is independent. For example: x + y = 5 and  - x + 2 y = 10. When we add those equations: 3 y = 15; y = 15 : 3; y = 5; x + 5 = 5; x = 0. So we have just one solution ( x , y ) = ( 0, 5 ). The system of equations is independent.
7 0
4 years ago
What’s the differential equation of
worty [1.4K]

Answer:

y=\frac{1}{x^{2}-6x+13 }

Step-by-step explanation:

We have given,

                        \frac{dy}{dx}=y^{2}(6-2x)

and initial condition x=3,\  y=\frac{1}{4}

Now,

\frac{dy}{dx}=y^{2}(6-2x)

Rearranging the variables, we get

\frac{dy}{y^{2} }=(6-2x).dx

Applying integration both sides, we get

\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx

⇒\frac{-1}{y} =6x-\frac{2x^{2} }{2}

⇒ \frac{-1}{y}=6x-x^{2}  +C                  

Putting the initial condition (i.e., x=3,\  y=\frac{1}{4}), we get

⇒ -4=6\times3-(3)^{2}+C

⇒ -4=18-9+C

∴ C=-13

We have,  \frac{-1}{y}=6x-x^{2}  +C    

now putting the value of C in above equation, we get

⇒ \frac{-1}{y}=6x-x^{2}  -13

⇒  \frac{1}{y}=-6x+x^{2}  +13

y=\frac{1}{x^{2}-6x+13 }

5 0
3 years ago
On Monday it was 70 degrees. On Tuesday it dropped to 45 degrees. What was the percent of change?
Tomtit [17]

Answer:

The answer is 64.29% hope it helps

6 0
3 years ago
Read 2 more answers
Can u guys PLEASE ANSWER this question 2 ASAP
kicyunya [14]

Answer:

Sorry but where's the question

Step-by-step explanation:

6 0
3 years ago
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