To solve for x in a triangle with side lengths of 67, 29, and x, we can use the Law of Cosines. The value of x is approximately 47.6.
To solve for x in the given triangle with side lengths 67, 29, and x, we can use the Law of Cosines. The Law of Cosines states that for any triangle with side lengths a, b, and c and angle C opposite side c, the following equation holds true: c^2 = a^2 + b^2 - 2ab*cos(C). In this case, we can substitute the given values into the equation and solve for x. Let's calculate it:
x^2 = 67^2 + 29^2 - 2*67*29*cos(C)
x^2 = 4489 + 841 - 3886cos(C)
Solving for x, we find that x is approximately 47.6.
#SPJ12
Answer:
84°
Step-by-step explanation:
Because every triangle has a combined side length of 180°
67+29=96°
180-96=84°
Answer:
39
Step-by-step explanation:
27 - -12 is 39, friend.
Answer:
39
Step-by-step explanation:
Answer:
Around 9.57 minutes to edit
Step-by-step explanation:
Do 28 divided by 6 to get around 4.7. Then do 45 divided by 4.7 to get the answer.
Answer:
the missing length is 5
Step-by-step explanation:
Since we are given similar triangles, we can simply use proportions to solve the problem:
we say that 2 is to side of length 4, the same as m is to side lengths 10. And in math terms:
5x - 7y=40
O (1.-5)
O (1,5)
0 (-1,-5)
0 (-1,5)
Answer:
x=1 ,y=-5
Step-by-step explanation:
3x + 10y = 47
5x - 7y = 40
____________
21x + 70y = 329. equ 1
50x - 70y = 400. equ 2
subtract equ 1 from 2
=71x = 71
divide both sides by the 71
71/71 = 71/71
x = 1
_________
to find y
substitute x into either equ 1 or equ 2
am using equ 2
5x - 7y = 40
5(1) - 7y = 40
5 - 7y = 40
7y = 5 - 40
7y = -35
divide both sides by the coffecient of y
7y/7 = -35/7
y = - 5
Answer:
1) h = -1/2t^2 +10t
2) h = -1/2(t -10)^2 +72
3) domain: [0, 20]; range: [0, 50]
Step-by-step explanation:
1.) I find it easiest to start with the vertex form when the vertex is given. The equation of the presumed parabolic path for Firework 1 is ...
h = a(t -10)^2 +50
To find the value of "a", we must use another point on the graph. (0, 0) works nicely:
0 = a(0 -10)^2 +50
-100a = 50 . . . . . . subtract 100a
a = -1/2 . . . . . . . . . divide by -100
Then the standard-form equation is ...
h = (-1/2)(t^2 -20t +100) +50
h = -1/2t^2 +10t
__
2.) The path of Firework 2 is translated upward by 22 units from that of Firework 1.
h = -1/2(t -10)^2 +72
__
3.) The horizontal extent of the graph for Firework 1 is ...
domain: 0 ≤ t ≤ 20
The vertical extent of the graph for Firework 1 is ...
range: 0 ≤ h ≤ 50