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Naddik [55]
3 years ago
13

2/x-1 + 8/2x-2 = 7 solve the equation (the answer has to be a fraction)

Mathematics
1 answer:
bogdanovich [222]3 years ago
3 0
\frac{2}{x - 1} + \frac{8}{2x - 2} = 7
\frac{2}{x - 1} + \frac{2(4)}{2(x) - 2(1)} = 7
\frac{2}{x - 1} + \frac{2(4)}{2(x - 1)} = 7
\frac{2}{x - 1} + \frac{4}{x - 1} = 7
\frac{2 + 4}{x - 1} = 7
\frac{6}{x - 1} = 7
7(x - 1) = 6
7(x) - 7(1) = 6
7x - 7 = 6
7x = 13
\frac{7x}{7} = \frac{13}{7}
x = 1\frac{6}{7}
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Marshall drove to work in rush hour traffic, going just 27 miles in 1 hour and 20 minutes. Later that day, he left work early to
iragen [17]

Answer:

The average speed of Marshall is 29.7 mph

Step-by-step explanation:

Given;

distance traveled to work, d₁ = 27 miles

distance traveled back home, d₂ = 27 miles

initial time taken, t₁ = 1 hour and 20 minutes

final time taken, t₂ = 29 minutes

Total distance traveled, d = d₁ + d₂ = 27 miles + 27 miles = 54 miles

Total time taken for the journey, t = t₁ + t₂ = 1 hour + ( 20 mins + 29 mins)

t = 1 hour + 49 mins = 1 hour + 0.817 hour = 1.817 hour

The average speed is given by;

v = total distance / total time

v = 54 miles / 1.817 hour

v = 29.7 mph

Therefore, the average speed of Marshall is 29.7 mph

8 0
3 years ago
What is -x-3y=-3 and 2x+3y=5 using elimination method?
xxTIMURxx [149]

Answer:(2,1/3)


Step-by-step explanation:

-x-3y=-3

2x+3y=5


_________First solve for x since y is 0. -3y+3y=0

1X = 2

1x/1= 2/1

X= 2

Now substitute your x answer into any equation to find y I used second equation

2x+3y= 5

2(2)+3y=5

4+3y=5

3y=5-4

3y=1

3y/3=1/3

Y=1/3



6 0
4 years ago
Find a set of vectors {u⃗ ,v⃗ }{u→,v→} in r4r4 that spans the solution set of the equations {w−x+2y+4z5w+2x−y+3z==0,0. {w−x+2y+4
rusak2 [61]
Given

w-x+2y+4z=0 \\ 5w+2x-y+3z=0

We can rewrite it in matrix form as:

\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right]   \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2
\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \  \frac{1}{7} R_2\rightarrow R_2 \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1
\\  \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\  \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \

Thus, the solution set is a span of \{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}
7 0
3 years ago
Find the amount and compound interest on 16000 for 2 years at 15%, interest being
Nat2105 [25]

Answer:

Step-bAmount at the end of 2 years = Rs. 21,160

Amount at the end of first year = Rs. 18,400

So interest for first year = Rs. 2400

Interest for second year = Rs. 2760

Explanation:

A = P (1 + r/n)^nt

P = 16,000

A = ?

r = 15% = 15/100

n= 1

t = 2

Substituting the values in the formula, we get:

A = 16000 (1 + 15 / 100) ^2

A = 16000 (23 / 20) ^2

A = 16000 * 23 * 23 / 400

A = 40 * 23 * 23

A = 21160

Amount at the end of 2 years = Rs. 21,160

Amount at the end of first year = Rs. 18,400

So interest for first year = Rs. 2400

Interest for second year = Rs. 2760

hope helpful

5 0
3 years ago
tickets to the concert were 2.50$ for adults and 1$ for students 1200$ was collected and 750 tickets were sold
Lelechka [254]

450 student tickets were sold and 300 adult tickets were sold (I believe this is what was meant to be asked.

It won’t let me add attachments without reloading and not working, so if you need the work just ask and I will be happy to type it out.

4 0
4 years ago
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