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oee [108]
3 years ago
5

Help I need this done soon!!! Thank you!!

Mathematics
1 answer:
jenyasd209 [6]3 years ago
4 0
D is the midpoint of segment AC, Given
Segment AD is congruent to segment CD, Definition of Midpoint
Angle BDC is congruent to angle BDA, Given
Segment BD is congrent to angle BD, Reflexive property of congruence
Triangle ABD is congruent to triangle CBD, SAS
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Find the difference between 1.97 and 20
cestrela7 [59]

Answer:

1.97-20= -18.03

20-1.97= 18.03

Step-by-step explanation:subtract the numbers from each other.

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3 years ago
How much is 13,200 feet in miles
Naddika [18.5K]

Answer:

2.5 miles

Step-by-step explanation:

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3 years ago
A. Find the reflection of the point (-3, 2) about the origin.
nadezda [96]

Answer:

a. (3,-2)

b. (-5,-3)

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3 years ago
What is the difference between the predicted value and the actual value
snow_lady [41]
It’s a prediction so basically an estimate of how much students he thought were going to be there. An actual number is the real total.
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3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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