The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
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Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
PART A:
The table of h(t) and g(t) for the value of x between 4 and 7 inclusive are shown below.
The solution for h(x)=g(x) lies between x = 4 and x = 5, this is becuase the range of value given by h(4) and h(5) and g(4) and g(5).
The graph is shown in picture 2 to confirm this. The two functions intersect each other between x = 4 and x = 5
PART B:
The point of intersection is the point where the cannon balls will collide.
Answer:
The answer is (A.)
Step-by-step explanation:
4 x 9 = 36
6 x 9 = 54
8 x 9 = 72
Answer:
C) 6 feet
Step-by-step explanation:
set up a proportion of height/shadow = height/shadow
30/10 = h/2
cross-multiply:
10h = 60
h = 6
Answer:
c.
Step-by-step explanation:
The X value 1 is repeated
