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Tomtit [17]
3 years ago
11

Hola me ayudan ???????????

Mathematics
1 answer:
prisoha [69]3 years ago
3 0

Answer:

AAAAAA

Step-by-step explanation:

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A triangle has sides with the lengths and angle measures given. Classify each triangle. Write scalene, isosceles, or equilateral
djyliett [7]
1) scalene; obtuse
2)scalene; acute
7 0
4 years ago
Read 2 more answers
1. Which function represents exponential decay?
MatroZZZ [7]

Answer:

y= 7^-t

Step-by-step explanation:

Equation for exponential decay:

An equation represents exponential decay if a term between 0 and 1 is elevated to the independent variable.

In this question:

First option 4 is elevated, so not exponential decay.

Second option:

7^{-t} = \frac{1}{7^t} = (\frac{1}{7})^{t}

0 < \frac{1}{7} < 1, thus, the second option represents exponential decay.

7 0
3 years ago
Please help I'm almost done with this course I just need a few more answered
iogann1982 [59]

The answer would be B.

Hope this helped :)

6 0
3 years ago
Please I need answer right now ppleassssee<br><br>D. -1/4+-2/3<br>with explanation pleaseeee​
Masja [62]

Answer:

1 - 1/3

Step-by-step explanation:

Since 1/2 * 2/3 = 1/3, it should be the case that 1/3 divided by 2/3 gives 1/2 and that 1/3 divided by 1/2 gives 2/3. To divide fractions, we multiply the numerator by the reciprocal of the denominator, where the reciprocal of a number just interchanges the numerator and denominator of the number.

5 0
3 years ago
How do you do this trig problem.
loris [4]

Consider a right triangle in which one of the angles \theta satisfies

\sin\theta=\dfrac35\implies\theta=\sin^{-1}\dfrac35

That is, the side opposite \theta occurs in a ratio of 3 to 5 with the hypotenuse. The side adjacent to \theta then occurs in a ratio of 4 to 5 with the hypotenuse. In other words,

\cos\theta=\dfrac45

because

\sin^2\theta+\cos^2\theta=\dfrac9{25}+\dfrac{16}{25}=1

Then in this triangle,

\cot\theta=\cot\left(\sin^{-1}\dfrac35\right)=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac45}{\frac35}=\dfrac43

7 0
3 years ago
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