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OLEGan [10]
3 years ago
9

A park plants young maple trees. A maple tree grows 1 1/2 feet per year.

Mathematics
2 answers:
zzz [600]3 years ago
6 0
It'll be 30 feet tall and it'll take 40 years to grow to it's full height.
nekit [7.7K]3 years ago
3 0
I think you do multiplying fractions. So 1 1/2 x 20
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Y + (-3 y2 ) +2( y2 - 6y) simplify​
labwork [276]

Answer:

y - 3y^2 + 4y^2 - 12y

y^2 - 11y

Step-by-step explanation:

3 0
3 years ago
Two teams are distributing information booklets. Team A distributes 60% more boxes of booklets than Team B, but each box of Team
Alexxx [7]

Answer:

  C.  4,100

Step-by-step explanation:

"60% more" is represented by a multiplier of 1 + 0.60 = 1.60.

"60% fewer" is represented by a multiplier of 1 - 0.60 = 0.40.

__

Let b represent the number of booklets distributed by Team B. Then the number distributed by Team A is ...

  1.60 × (0.40b) . . . . 60% more boxes, each with 60% fewer booklets

  = 0.64b

Then the total distributed by both teams is ...

  b + 0.64b = 1.64b = (164/100)b = (41/25)b

The only answer choice that is a multiple of 41 is ...

  4,100 . . . choice C

__

For 4100 to be the number of booklets distributed by both teams, Team B will have distributed 2500 booklets, and Team A will have distributed 1600 booklets. Team A might have distributed 160 boxes of 100 booklets, while Team B might have distributed 100 boxes of 250 booklets.

7 0
3 years ago
Explain one strAtegy for finding 2 2/5 + 1 2/5.
Monica [59]

Answer:

3 4/5

Step-by-step explanation:

just add the whole number then add the numerator then the denominator stay the same

4 0
3 years ago
Read 2 more answers
What is the value of a in the equation 5a-10b=45 when b=3
valentina_108 [34]
Since b = 3 the equation changes

5a - 10(3) = 45
5a - 30 = 45

(Add 30 to both sides)
5a = 75
Divide 5 to both sides
a = 15
7 0
3 years ago
Read 2 more answers
If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

6 0
3 years ago
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