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stich3 [128]
2 years ago
7

Couldn’t figure this out help please

Mathematics
2 answers:
Bogdan [553]2 years ago
5 0

9514 1404 393

Answer:

  B. x + y < 0

Step-by-step explanation:

The two equations can be cleared of fractions by multiplying by 15.

  15(2/3(x +1) -4/5y) = 15(1/3)

  10(x +1) -12y = 5

  10x -12y = -5

and

  15(2/5x +1/3(2y +1)) = 15(1/5)

  6x +5(2y +1) = 3

  6x +10y = -2

  3x +5y = -1 . . . . . eliminate common factor of 2

__

You can find the solutions any way you like, but you can answer the question without doing that. The lines are not parallel, nor coincident, so there is exactly one solution. (choices C and D are incorrect)

If we can locate the solution relative to the line x + y = 0, we can tell if choice A or choice B is correct. A quick look at the intercepts of the equations tells us the solution cannot lie in quadrants 1 or 4. The negative y-intercept and shallow slope (-3/5) of the second equation tells us the solution must lie below the line x + y = 0. That means x+y < 0, choice B.

_____

In the attached graph, the line x+y=0 is dashed orange. Above that line, x+y>0; below that line, x+y<0. We see the intersection point of the red and blue lines is in the region where x+y < 0.

For standard form equation ax+by = c, the x- and y-intercepts are c/a and c/b, respectively, so are easy to find from that form. Knowing these makes it easy to make a sketch of the graph, locating the solution point relative to the line x+y = 0.

WARRIOR [948]2 years ago
4 0

(B)

Step-by-step explanation:

Rewrite the equations into their standard forms. The first one can be rewritten as

10x - 12y = -5

and the 2nd can be rewritten as

3x + 5y = -1

Solving this system either by substitution or elimination, we get

x = -\dfrac{37}{86}\:\:\text{and}\:\:y= \dfrac{25}{86}

If you add x + y, you'll get a negative number.

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Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
Ivan rented a truck for the day. There is a base fee of $17.95, and there was an additional charge of 91 cents for each mile dri
lozanna [386]

Answer:

293 miles

Step-by-step explanation:

284.58-17.95=266.63

266.63/0.91=293

293 miles

7 0
2 years ago
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