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Elenna [48]
3 years ago
11

Dolbear’s law states the relationship between the rate at which snow a tree crickets chirp and the air temperature of their envi

ronment the formula is T = 50+ N - 40 over 4 where T is a temperature in degrees Fahrenheit and N is a number of chirps per minute if N = 66 find the temperature in degrees Fahrenheit. T
Mathematics
1 answer:
STatiana [176]3 years ago
8 0

Answer:

T=76

Step-by-step explanation:

If the formula is T=50+N-40 and N=66 then you just substitute 66 in for N then you have T=50+66-40 where as T=76

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Which arrangement shows 275 , 5.36, 535 , and 336 in order from least to greatest
CaHeK987 [17]
The answer would be B.
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2 years ago
What is the GCF of 35 and 49??
Vlada [557]

Answer: 7

Step-by-step explanation: 7x7=49 and 7x5=35

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Which expression is equivalent to 200?<br> O 2 /10<br> O 10/2<br> O 10/20<br> O 100/13
creativ13 [48]

Answer:

10/20

Step-by-step explanation:

10 times 20 is 200 which would give you 10/20

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2 years ago
What is mathematical equation?​
Talja [164]

Answer:

A statement asserting the equality of two expressions, usually written as a linear array of symbols that are separated into left and right sides and joined by an equal sign.

Step-by-step explanation:

A statement asserting the equality of two expressions, usually written as a linear array of symbols that are separated into left and right sides and joined by an equal sign.

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2 years ago
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Find all solutions of each equation on the interval 0 ≤ x &lt; 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

u^{2} + 1 = 2 u^{2}.

u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
2 years ago
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