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Dennis_Churaev [7]

2 years ago

6

Brett writes a number that is the same distance from o as 6 on a number line but in the opposite direction. What number did bret

t write

1 answer:

photoshop1234 [79]2 years ago

4 0

Well since 6 would be in the oppposite direction it would be turned around. Well this leads to non than any other number 9.

Answer ~ 9

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A nick on the edge of a CD rotates to (−6, 5) during one song when represented graphically.

Allisa [31]

By definition, the sine of an angle is given by:
$sin \alpha = \frac{C.O}{h}$
Where,
C.O: opposite leg
h: hypotenuse
The hypotenuse is given by:
$h = \sqrt{(-6)^2+(5)^2}$
$h = \sqrt{(61}$
So, replacing values we have:
$sin \alpha = \frac{5}{\sqrt{61}}$
Rewriting we have:
$sin \alpha = \frac{5\sqrt{61}}{61}$
Answer:
the sine value of this function is:
A)5 times square root 61 over 61

6 0

3 years ago

Read 2 more answers

A new gym member wants to burn 525 calories during the first workout and prefers the stationary bike but is considering doing ae

S_A_V [24]

Sounds like you’re missing a piece in this problem

7 0

3 years ago

Solve for the given variable 3(2x+5)=4x+21

Nookie1986 [14]

Answer:

x=3

Step-by-step explanation:

3(2x+5)=4x+21

6x+15=4x+21

-4x -4x

2x+15= 21

-15 -15

2x=6

2 2

<h2>x=3 Hope that helped!!!!</h2>

6 0

3 years ago

Help me pls only 10points

NeX [460]

y=-7x+8

y=-7x-8

-7x+8=-7x-8

8=-8

x∈∅

B. No solutions

3 0

3 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

1 year ago