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Naddika [18.5K]
3 years ago
8

Item 19 A plant has an initial height of 1 inch and grows at a constant rate of 3 inches each month. A second plant that also gr

ows at a constant rate has an initial height of 4 inches and is 28 inches tall after 1 year. After how many months are the plants the same height?
Mathematics
1 answer:
Alex777 [14]3 years ago
3 0
3 months, i just answered this same question lol.
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Write 18/31 in simplest form. Use / for fraction line.​
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Answer: 18/31

It is already the simplest form, so it cannot be simplified.
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2 years ago
The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func
katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle (\frac{1}{3},\frac{1}4).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A(\frac{1}{3},\frac{1}4)

Let O be the origin i.e. (0,0)

Point B will be (\frac{1}{3},0)

Now, let us consider the right angled triangle \triangle OBA:

Sides:

Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}

Using Pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}

sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}

\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}

cos\angle AOB = \dfrac{Base}{Hypotenuse}

\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}

tan\angle AOB = \dfrac{Perpendicular}{Base}

\Rightarrow tan\angle AOB = \dfrac{3}{4}

cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}

\Rightarrow cosec\angle AOB = \dfrac{5}{3}

sec\angle AOB = \dfrac{Hypotenuse}{Base}

\Rightarrow sec\angle AOB = \dfrac{5}{4}

cot\angle AOB = \dfrac{Base}{Perpendicular}

\Rightarrow cot\angle AOB = \dfrac{4}{3}

3 0
2 years ago
_____ appear to never cross or meet and are always the same distance apart
Akimi4 [234]

Answer: parallel lines

Step-by-step explanation:

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I think it d. I’m not sure I could be wrong
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Answer:

The answer is B.

Step-by-step explanation:

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