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tigry1 [53]
1 year ago
12

There are 25 popular trees currently in the park. Park workers will plant morepopular trees today. When the workers are finished

there will be 80popular trees in the park. How many popular trees did the workers plant today?
Mathematics
1 answer:
Vlada [557]1 year ago
7 0

Current trees = 25

trees when workers are finished = 80

Subtract the number of current trees (25) to the number of trees that are when the workers are finished:

80-25 = 55

The workers planted 55 trees today

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Last summer, Marion was 3 1/2 feet tall. She was 4 inches taller than her brother Elijah. She was 1 1/4 feet shorter than her si
solniwko [45]

Answer:

Answers below

Step-by-step explanation:

Marion: 42 inches

Elijah: 38 inches

Lorie: 57 inches

Convert each to inches and solve.

Hope this helps!

8 0
3 years ago
The difference of 3 times a number and 20 is greater then 10
Novosadov [1.4K]
If the number is x, then 3 times x -20 >10 and 3x-20>10. Adding 20 to both sides, we get 3x >30 and by dividing by 3 we get x >10
3 0
3 years ago
Predict:<br> y=20<br> x=35<br> Predict for x=40
11Alexandr11 [23.1K]
X is left and right

hmm
see the aproximate slope
see endpoints
ok, so from about (0,3) to (35,30)
slpoe is 27/35

so
y=(27/35)x+3 about
so when x=
ok, so according to the garph, when x=35, y=30, not 20


ok, so when x=40

y=(27/35)x+3
y=(27/35)*40+3
y=33.8571 about for x=40
6 0
3 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
The ordered pair, (1,5) indicates the unit rate of books to cost on the graph shown. What does the point on the graph represent?
DerKrebs [107]
The point on the graph represents that 2 books cost $10.
3 0
2 years ago
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