x = 4 and EF = 4 ⇒ 3rd answer
Step-by-step explanation:
If two triangles are congruent, then
1. Their corresponding sides are equal
2. Their corresponding angles are equal
3. Their areas and perimeters are equal
∵ △ ABC ≅ △ DEF
∴ AB ≅ DE
∴ BC ≅ EF
∴ AC ≅ DF
∵ BC = 4 x - 12
∵ EF = -3 x + 16
∵ BC = EF
∴ 4 x - 12 = -3 x + 16
Let us solve the equation to find x
∵ 4 x - 12 = -3 x + 16
- Add 3 x for both sides
∴ 7 x - 12 = 16
- Add 12 to both sides
∴ 7 x = 28
- Divide both sides by 7
∴ x = 4
Substitute the value of x in the expression of EF
∵ EF = -3 x + 16
∵ x = 4
∴ EF = -3(4) + 16 = -12 + 16
∴ EF = 4
x = 4 and EF = 4
Learn more:
You can learn more about congruence of Δs in brainly.com/question/3202836
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Answer:
Equation is: y = 0.5x² + 0.5x - 3
Explanation:
general form of the parabola is:
y = ax² + bx + c
Now, we will need to solve for a, b and c.
To do this, we will simply get points from the graph, substitute in the general equation and solve for the missing coefficients.
First point that we will use is (0,-3).
y = y = ax² + bx + c
-3 = a(0)² + b(0) + c
c = -3
The equation now becomes:
y = ax² + bx - 3
The second point that we will use is (2,0):
y = ax² + bx - 3
0 = a(2)² + b(2) - 3
0 = 4a + 2b -3
4a + 2b = 3
This means that:
2b = 3 - 4a
b = 1.5 - 2a ...........> I
The third point that we will use is (-3,0):
y = ax² + bx - 3
0 = a(-3)² + b(-3) - 3
0 = 9a - 3b - 3
9a - 3b = 3 ...........> II
Substitute with I in II and solve for a as follows:
9a - 3b = 3
9a - 3(1.5 - 2a) = 3
9a - 4.5 + 6a = 3
15a = 7.5
a = 7.5 / 15
a = 0.5
Substitute with the value of a in equation I to get b as follows:
b = 1.5 - 2a
b = 1.5 - 2(0.5)
b = 0.5
Substitute with a and b in the equation as follows:
y = 0.5x² + 0.5x - 3
Hope this helps :)
Answer:
Coldest has - 2° C while warmest had 2 °C
Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)
The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)
At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots
The distance between the ships is increasing at about 28.37 knots at 4 pm.
