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.
Answer:
I cant see it
Step-by-step explanation:
We need to write an expression for "add 34 and 6 and then multiply 3"
"Add" represents + operation .
"multiply" represents × operation.
We need to add 34 and 6.
So, we can write 34+6 and then we need multiply the sum by 3.
So, we need to keep above addition of 34 and 6 in parantehsis. We get
(34+6) × 3.
So, the final expression is (34+6) × 3.
Answer:
No solutions I think
Step-by-step explanation:
The answer to your problem is 108.18
