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butalik [34]
3 years ago
13

Julia evaluated the expression 3 to the third power plus 20 divided by 2 to the second power. What’s the answer need it tonight

Mathematics
1 answer:
nexus9112 [7]3 years ago
8 0

Answer:

32

Step-by-step explanation:

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The points (6,y) and (7,9) fall on a line with a slope of 5 . ​ ​What is the value of y ? ​
JulsSmile [24]

Answer:

y=4 (6,4)

Step-by-step explanation:

hope this helps!

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3 years ago
Liability that arises from not maintaining a building is referred to as a) medical foundation liability. b) premises liability.
Sauron [17]

Answer:

opt. B

Step-by-step explanation:

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Math 9
Nata [24]
The answer is C (-3, -5), D(1,-5)
4 0
2 years ago
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A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each
Mashcka [7]

Answer: There are 30 participants are needed for the entire study.

Step-by-step explanation:

Since we have given that

Number of levels of factor A = 2

Number of levels of factor B = 3

Number of participants in each treatment condition = 5

So, the number of participants are needed for the entire study is given by

2\times 3\times 5\\\\=6\times 5\\\\=30

Hence, there are 30 participants are needed for the entire study.

7 0
3 years ago
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist
yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0
3 years ago
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