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goblinko [34]
3 years ago
13

What are the five congruence postulates and theorems we have learned to prove that two triangles are congruent? Which of these r

easons proves that triangle ABC is congruent to triangle CDA? How did you arrive at your answer? Be sure to answer in complete sentences.

Mathematics
2 answers:
tangare [24]3 years ago
5 0
Angle side Angle
parallel lines cut by a transversal form alternate interior angles 
parallel lines are congruent 
opposite angles are congruent ( angle D and angle B) 
line AB is congruent to line BC 

Sauron [17]3 years ago
4 0
The correct answer is ASA
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A growth factor is a value that is greater than 1?
amm1812

Answer:

No, it is a value that is greater than 0

Step-by-step explanation:

A growth factor is indeed a value greater than 1 except it can be a decimal below 1 to. Therefore, a growth value is any POSITIVE number.

6 0
2 years ago
Question 6 options:<br><br> tan D<br><br><br> tan F<br><br><br> cos F<br><br><br> sin F
ivanzaharov [21]

Answer:

tan F

Step-by-step explanation:

tangent is opposite/adjacent

the opposite side of F is 8 and the adjacent is 15

8/15 is therfore = tan F

7 0
2 years ago
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Solve the following system. Please show me how you did it.
Andrej [43]

Answer:

x = 3, y = 4 and z = -5.

Step-by-step explanation:

Use the elimination method:

-3x + 2y + z = -6...........(1)

x + 3y + 2z = 5..............(2)

4x + 4y + 3z = 13..........(3)

Multiply the first equation by -2:

6x - 4y - 2z = 12   Add this to equation (2):

7x  - y  = 17...........(4)

Now Multiply equation (2) by 3 and equation (3) by -2

3x + 9y + 6z = 15 ............(5)

-8x - 8y - 6z = -26............(6)   Adding (5) + (6):

-5x + y = -11..........(7)

Now Add (7) and (4):

2x = 6

x = 3.

Now  substitute for x in (4):

7(3) - y = 17

-y = 17 - 21 = -4

y = 4.

Finally substitute for x and y in equation (3) to find z:

4(3) + 4(4)  +3z = 13

3z = 13 - 12 - 16 = -15

z = -5.

4 0
2 years ago
(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in s
Bezzdna [24]

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

5 0
2 years ago
A teacher wants to know whether the students in the entire school prefer watching television or playing outdoor games after scho
cupoosta [38]
D. All students in each grade. It can't be A since the survey is for the students, it defeats the purpose. It can't be B and C since those are too biased. Basing the student population on one factor like gender and sports is not logical in the context of this survey. 
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2 years ago
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