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Anni [7]
3 years ago
7

Divide a salad with six friends, 1/3 is missing, how much does each person get?

Mathematics
1 answer:
Talja [164]3 years ago
4 0
Each friend gets 1/9. If you have 1 salad and 1/3 is missing that leaves you with 2/3, you have 6 friends and I am assuming they each get an equal amount so 2/3 equals 4/6, 4/6 equals 6/9, there are 6 people, they each get 1/9.
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Determine if the lines are parallel, perpendicular or neither y=-3/2x-4 and y=2/3x-7
aliya0001 [1]

Answer:

<u>They are perpendicular</u>

Step-by-step explanation:

Perpendicular lines have slopes that are negated versions of the other or positive versions of the slope of the other. The slopes are also flipped.

Parallel lines have the same slope.

Neither is when the lines do not follow either of these definitions.

The slope of your two lines are -3/2 and 2/3. These slopes are flipped and the opposite negations of each other. <u>This means the lines are perpendicular.</u>

<u />

4 0
3 years ago
What is the slope of the line with these points (5, -3) (6, -1) (7, 1)
ycow [4]

Answer:

4/2

Step-by-step explanation:

Rise/Run

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8 0
3 years ago
Read 2 more answers
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

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