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sertanlavr [38]
3 years ago
7

Given the following system of equations, identify the type of system.

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
8 0

Answer:

the answer would be B inconsistent

ICE Princess25 [194]3 years ago
7 0

Answer:

The two lines in the graph are parallel, so it is B. Inconsistent since there is no solution.

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A consistent, dependent system of equations is a system with __________. (1 point)
Lesechka [4]

Answer:  hope this helps

Step-by-step explanation:  intersecting lines that have the same slope and the same y-intercept

6 0
3 years ago
How do i write 1.67 in words
Alex Ar [27]
Many people would say "one point six seven" or "one dot six seven".

But technically, it's "<em>one and sixty-seven hundredths</em>".
5 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

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

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(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

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Jessica just put 14.41 gallons of fuel into her car. There were 3.6 gallons in the car to begin with. How much fuel is in jessic
SIZIF [17.4K]

Answer:

18.01 gallons

Step-by-step explanation:

14.41

3.6

=18.01

8 0
3 years ago
Let f(t) give the number of liters of fuel oil burned in t days, and w(r) the liters burned in r weeks. Find a formula for w by
viktelen [127]

Answer:

7 f(t)

Step-by-step explanation:

So, our f(t) is the number of liters burned in t days. If t is 1, f(t)=f(1) and so on for every t.

w(r) id the number of liters in r weeks. This is, in one week there are w(1) liters burned.

As in one week there are 7 days, we can replace the r, that is a week, by something that represents 7 days. As 1 day is represented by t, one week can be 7t (in other words r = 7t). So, we have that the liters burned in one week are:

w(r) = w[7f(t)]

So, we represented the liters in one week by it measure of days.

So, we can post that the number of liters burned in 7 days is the same as the number of liters burned 1 day multiplied by 7 times. So:

w (r) = w[7 f(t)] = 7 f(t)

Here we hace the w function represented in terms of t instead of r.

3 0
3 years ago
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