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GREYUIT [131]
3 years ago
9

A system of linear equations is graphed. Which ordered pair is the best estimate for the solution to the system? (12, 0) (1, 1)

(12, 1) (13, 0) A system of 2 linear equations graphed on a coordinate plane. The horizontal x-axis ranges from negative 5 to 5 in increments of 1. The vertical y-axis ranges from negative 5 to 5 in increments of 1. First line passes through begin ordered pair 0 comma 2 end ordered pair and begin ordered pair 1 comma negative 1 end ordered pair. Second line passes through begin ordered pair 0 comma negative 1 end ordered pair and begin ordered pair 2 comma 2 end ordered pair.

Mathematics
2 answers:
omeli [17]3 years ago
7 0

WHELP.

Here's you're answer.

GuDViN [60]3 years ago
6 0
Line 1--------- >(0,2)     (1,-1)
y=mx+b         m=(-1-2)/(1-0)=-3
2=(-3)*(0)+b----- > b=2
y1=-3x+2

Line 2--------- >(0,-1)     (2,2)
y=mx+b         m=(2+1)/(2-0)=3/2
-1=(3/2)*(0)+b----- > b=-1
y2=(3/2)x-1

<span>using a graphic tool
see attached figure</span>

the best estimate solution for the system is <span>(1, 1) 
because </span>is the only one within the range of the x axis (-5,5) and the y axis (-5,5)

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8 0
3 years ago
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
Please help me I don’t have much time please look at the picture
Andre45 [30]

Answer:

Just add the painted miles

2 1/6 + 1 2/3

=3 5/6

then we subtact the total miles that wanna be painted which is 6

6 - 3 5/6

=2 1/6

The answer B

4 0
3 years ago
Read 2 more answers
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