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4 years ago

Point Q is located at (-4, 6) Point R is located at (8, 6) What is the distance from point Q to point R?

2 answers:

8 0

$\text {Distance = } \sqrt{(Y2-Y1)^2+ (X2-X1)^2}$

$\text {Distance = } \sqrt{(6-6)^2+ (8+4)^2}$

$\text {Distance = } \sqrt{12^2}$

$\text {Distance = } 12 \text { units}$

Answer: Distance between QR = 12 units.

Eddi Din [679]4 years ago

7 0

$\sqrt{ (8+4)^{2}+ (6-6)^{2} }$
↓
$\sqrt{ 12^{2}+ 0^{2} }$
↓
$\sqrt{144+0}$
↓
$\sqrt{144}$
↓
12
The distance between Points $Q$ and $R$ is 12 units.
Have a nice day!

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4 years ago

How do you do (a) and (b)?

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Answer:

See solution below

Step-by-step explanation:

(a) If n=0 or 1, the equation

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reordering:

$(1-n)y^{-n}y' = (1-n)y^{-n}ya(t)+(1-n)y^{-n}y^nf(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)y^{0}f(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)f(t)$

Now, using (2) and (3) we get:

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(b)

The equation we want to solve is

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$y' = -\frac{1}{x}y+x^3 y^3$

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$z(x)=x^2\int (\frac{-2x^3}{x^2})dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2\frac{x^2}{2}+Cx^2=-x^4+Cx^2$

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3 years ago

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Y-intercept:

The same goes for the opposite coordinate: if we picture a point at where the line intercepts the y-axis, there will be a point at -10 on the y-axis. Since the point is directly at 0 on the x-axis (meaning that it doesn't go left nor right) we know that the x-coordinate will be 0.

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3 years ago

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3 years ago

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