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amid [387]
2 years ago
14

Solve for x in the triangle. Round your answer to the nearest tenth. 13 32°

Mathematics
1 answer:
zloy xaker [14]2 years ago
5 0

Answer:

6.9

Step-by-step explanation:

imagine a circle with the center being the point with the 32 degree angle, and with the radius 13.

then

x = sin(32)×13 = 6.9 (6.88895...)

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Write the equation of a line that meets all the following requirements:
LuckyWell [14K]

9514 1404 393

Answer:

  • x + y = 5
  • y = -x + 5

Step-by-step explanation:

We can start with the point-slope form of the equation for a line. To meet the given requirements, we can use a point of (5, 0) and a slope of -1. Then the equation in that form is ...

  y -0 = -1(x -5)

Simplifying gives the slope-intercept form:

  y = -x +5 . . . . . . . use the distributive property to eliminate parentheses

Adding x to both sides gives the standard form:

  x + y = 5

__

<em>Explanation</em>

We know the line has the required intercept and slope because we chose those values to put into the point-slope form. Conversion from one form to another made use of the rules of equality, the additive identity element (y-0=y), and the distributive property.

8 0
2 years ago
What is the rate of decay for the function f(x) =2(0.95)^x ( x is on top of the parentheses)
SIZIF [17.4K]

Answer:

The decay rate is 5%.

Step-by-step explanation:

Let a substance is decaying at the rate of r% per hour from the initial value of P for t hours, then the final value of the substance is given by the function  

f(t) = P(1 - \frac{r}{100})^{t} ........... (1)

Comparing this equation with the original equation given as

f(x) = 2(0.95)^{x} ............ (2) we get,

(1 - \frac{r}{100}) = 0.95

⇒ \frac{r}{100} = 0.05

⇒ r = 5%.

Therefore, the decay rate is 5%. (Answer)

7 0
3 years ago
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I'm assuming your answers are listed as 1 through 4, not decimals.
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Yes! It can!

The 12 meters will be the hypotenuse, the 10 meters can be a side and the 8 meters can be another side.

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Omg I’m so sorry I’m not giving you your answer but I need to comments so I can post my questions hope you get your answer!
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3 years ago
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