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seropon [69]
3 years ago
15

In ΔIJK, k = 7.6 cm, j = 7.5 cm and ∠J=116°. Find all possible values of ∠K, to the nearest 10th of a degree.

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

No Possible Triangles

Step-by-step explanation:

Deltamath

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You are at a restaurant. You are ordering a drink. You can choose small or large. Then you can choose between sprite, diet coke,
wlad13 [49]

Answer:

so in all it's 4.

Step-by-step explanation:

1: for the drink size.

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4 0
3 years ago
_______________________<br> can someone help me solve this question? Thank you.
Zinaida [17]

Answer:

-60

Step-by-step explanation:

Factor the problem out using FOIL.

The end result is: −60x^{2}−84x+9

The value of coefficent a (the number in front of the x^2) is -60.

8 0
3 years ago
What is the slope of the line?
meriva
The slope of the line can be defined as y = mx + b, where m is the slope
To find the slope, one will need to use the expression Δy/Δx (y final - y initial)/(x final - x initial)
For this problem, that equals 1/3

Now we will use the point slope form
(y - y initial) = m (x - x initial):

y + 1 = 1/3 (x - 2)

y = 1/3x - 5/3
7 0
3 years ago
Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to
eduard
The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079 
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
ln(0.8) = (-0.0079 * t) * ln(e)
-0.223 = -0.0079 * t
t = -0.223 / -0.0079
t = 28.23
t ≈ 28 years
8 0
3 years ago
Factor the following polynomial.<br><br> ( Correct and best explained answer gets brainliest. )
mamaluj [8]
Glad you asked this question! Over the years of working with my professor we learned that if phs= 5 (constant) times the j to the power of 2 you will get the formula or polynomialic equation. It quite fascinating and due to using his equation the answer in fact is "c". Cheers
4 0
3 years ago
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