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3 years ago

7

This is a repost since i forgot the photo on the last one help plz

2 answers:

Gemiola [76]3 years ago

7 0

15)
(2,3), (8,-3), (2,-8), (-4,-3)

16)
(1,1), (1,7), (7,1), (7,7)

tatuchka [14]3 years ago

4 0

Answer:

Sorry but I don't know

Step-by-step explanation:

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yra solves the equation as shown. - 3(x+2) = 9 1. - 3x - 6 = 9 2. - 3x = 15 3. x = - 5 The property Tyra used in line 1 was the:

Molodets [167]

Answer:

Distributive property says that:

$a\cdot(b+c) = a\cdot b + a\cdot c$

Addition property of equality:

if x = y then x+z = y+z

Division property of equality:

if x = y then $\frac{x}{z}=\frac{y}{z}$

As per the statement:

Given the equation: $-3(x+2)=9$

Tyra solve this equation as shown below:

1.

$-3x -6 = 9$ [Distributive property]

2.

$-3x = 15$ [Addition property of equality]

3.

$x = -5$ {Division property of equality]

Therefore, the property Tyra used in line 1 was : Distributive Property.

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2 years ago

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Subtraction and division are derived from which principal operations

melisa1 [442]

Subtraction and division are from the principal operations of PEMDAS. PEMDAS stands for parentheses, exponents, multiply and divide, add and subtract and thats it! Hope this helps and remember to mark as brainliest!

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2 years ago

There are 5 times as many yellow labs as terriers in the dog park. if there are a total of 18 dogs, how many dogs are terriers?

gtnhenbr [62]

Terriers = x
Labs = 5x
5x + x = 18
6x = 18
6x / 6 = 18/6
x = 3
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3 years ago

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Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

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3 years ago

Evaluate: 5-3 xwhen x=-7 Substitute-7 for x.

Mice21 [21]

The answer for this question is 26

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3 years ago