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TEA [102]
3 years ago
15

1-z/z-7 - 8z-3/7-z subtract

Mathematics
1 answer:
maxonik [38]3 years ago
3 0
The answer to your question above would be: 7z-2/z-7

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how do you solve y=x+4 and y=-2x+1 using the subsituation method and show work and please give me the correct answer
SIZIF [17.4K]

Answer:

(-1,3)

Step-by-step explanation:

so y=x+4 and y=-2x+1

obviously, y=y

so that means that x+4 is equal to -2x+1

let's use that

y=x+4,y=-2x+1

y=y

x+4=-2x+1

x+2x=1-4

3x=-3

x=-1

y=x+4

y=(-1)+4

y=3

(-1,3)

6 0
3 years ago
I just want to know if I’m right
Svetradugi [14.3K]

Answer:

correct

Step-by-step explanation:

supplementary is when 2 angles add up to 180

8 0
3 years ago
Read 2 more answers
What is 8/15 plus 2/5 equal to
anastassius [24]

We have to add 8/15 to 2/5.

To add two fractions we have to make them have the same denominator, so we have to convert 2/5 to have a denominator of 15. To do this, we have to multiply 2/5 by 3 and then we can add the fractions:

\begin{gathered} \frac{8}{15}+\frac{2}{5} \\ \frac{8}{15}+\frac{2}{5}\cdot\frac{3}{3} \\ \frac{8}{15}+\frac{6}{15} \\ \frac{14}{15} \end{gathered}

Answer: 14/15

5 0
1 year ago
–2.5p – 20 = 9p + 37.5
Elenna [48]

Answer:

p = - 5

Step-by-step explanation:

–2.5p – 20 = 9p + 37.5

combine like terms:

- 2.5p - 9p = 37.5 + 20

simplify:

- 11.5p = 57.5

p = 57.5 / -11.5

p = - 5

6 0
3 years ago
Read 2 more answers
Solve the following ODE's: c) y* - 9y' + 18y = t^2
Nastasia [14]

Answer:

y = C_1e^{3t}+C_2e^{6t} + \dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = C_1e^{3t}+C_2e^{6t}

now calculating P.I.

P.I. = \frac{t^2}{D^2 - 9D +18}

P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

hence the complete solution

y = C.F. + P.I.

y = C_1e^{3t}+C_2e^{6t} + \dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

7 0
3 years ago
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