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3 years ago

Use substitution to solve the system of equations.. PLEASE ANSWER AND EXPLAIN..asap

Mathematics

2 answers:

Leya [2.2K]3 years ago

4 0

Answer:

the answer to number one is (13.3,2)

the answer to number two is (2,-2)

Step-by-step explanation:

Nataliya [291]3 years ago

4 0

1. -5+y=-3
3x-8y=24

5+2=-3
3x40/3 -8x2 =24

-3=-3
24=24
(x,y) = (40/3, 2)

2. -3x-4y=2
x+2y=-2

-3(-2-2y) -4y=2
y=2

x=-2-2x(-2)
x=2

(x,y) (2.-2)

-3x2-4x(-2) =2
2+2 x(-2) =-2
2=2
-2=-2
(2,-2)
hope this helps:)

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What are the x and y-intercepts of the line described by the equation?<br><br> 3x−9y=10.8

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3 0

3 years ago

Please help me thank you I appreciate it

Volgvan

Answer:

68°

Step-by-step explanation:

PRQ = SRQ gives:

3x-8 = 2x+6

x = 6+8 = 14

So PRQ = SRQ = 34

and

PRS = PRQ+SRQ = 68

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The box and whisker plot represents the scores made by two different classes on the same test. By comparing the length of the bo

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3 years ago

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What is the value of x? 1/2(x-14)+11=1/2x-(x-4)

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3 years ago

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If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago