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Arlecino [84]
3 years ago
12

WHat is the best method for this system?

Mathematics
2 answers:
aksik [14]3 years ago
5 0

Answer:

Elimination method

Step-by-step explanation:

We have a choice to eliminate the y or x:

I chose x

So now we double the second equation to make the co efficient of x the same:

6x-5y = -3

6x+4y = 24

Now we subtract both equations to eliminate the x's:

6x-5y = -3
-     -       -

6x+4y = 24

-9y = -27

y = 3

Now we substitute this value into either equation 1 or 2:
I chose 2

3x + 2(3) =12

Simplify:

3x + 6 =12

Subtract 6 from both sides:

3x + 6 -6 = 12 -6

Simplify:

3x = 6

Divide both sides by 3:

3x÷3 = 6÷3

Simplify:

x = 2

MariettaO [177]3 years ago
3 0

Answer:

(2, 3 ) by the elimination method

Step-by-step explanation:

6x - 5y = - 3 → (1)

3x + 2y = 12 → (2)

multiplying (2) by - 2 and adding to (1) will eliminate x

- 6x - 4y = - 24 → (3)

add (1) and (3) term by term to eliminate x

0 - 9y = - 27

- 9y = - 27 ( divide both sides by - 9 )

y = 3

substitute y = 3 into either of the 2 equations and solve for x

substituting into (2)

3x + 2(3) = 12

3x + 6 = 12 ( subtract 6 from both sides )

3x = 6 ( divide both sides by 3 )

x = 2

solution is (2, 3 )

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I think the answer is b
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3 years ago
Four students were asked to write an expression which has terms that have a greatest common factor of 8c. The expressions provid
wel

Answer:

Wyatt wrote the correct expression

Step-by-step explanation:

Here we are to evaluate which of the students wrote an expression with the highest common factor being 8c

For Tanya; 32c + 16c; 16c( 2 + 1); This is wrong as the highest common factor is 16c

For Wyatt, we have 8c - 48cd; factorizing, we have 8c(1-6cd); This is correct

For Xavier we have 36c-24c; 8 is not a factor of 36, so it won’t work

For Yang 8c + 40; 8(c+ 5); what we are trying to look for is 8c and not 8 so this is wrong also

3 0
3 years ago
Read 2 more answers
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
What is the equation of the following graph in vertex form? PLEASE HELP ME!
viktelen [127]

Answer: Choice B

y = (x+3)^2 - 1

==================================================

Recall that vertex form is y = a(x-h)^2 + k with the vertex being (h,k)

From the given graph, it shows that (h,k) = (-3,1) is the vertex. So we plug this pair of values into the general function above to get

y = a(x-h)^2 + k

y = 1(x-(-3))^2 + (-1)

y = (x+3)^2 - 1

which is the final answer

note: a = 1 is the case for all four answer choices, so we don't have to worry about the value of 'a'

6 0
3 years ago
Please help me I need help with my test review I can’t figure out m<1 and m<2
aksik [14]

Answer:

m1=12 and m2=168

Step-by-step explanation:

m1 + m2 = 180 \: (straight \: angle) \\ then \\ x - 12 + 7x = 180 \\ 8x - 12 = 180 \\ 8x = 180  +  12 = 192 \\ x = 192\div 8 \\ x = 24\\ hence \\ m1 = 24 - 12 = 12 \\ m2 = 7 \times 24 = 168.

6 0
4 years ago
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