<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.
As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation
∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm
Option C
For each value of y, -2 is a solution of -21 = 6y - 9
<u>Solution:</u>
Given, equation is – 21 = 6y – 9
We have to find that whether given set of options can satisfy the above equation or not
Now, let us check one by one option
<em><u>Option A) </u></em>
Given option is -5
Let us substitute -5 in given equation
- 21 = 6(-5) – 9
- 21 = -30 – 9
- 21 = - 39
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option B)</u></em>
Given option is 3
- 21 = 6(3) – 9
- 21 = 18 – 9
- 21 = 9
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option C)</u></em>
Given option is -2
- 21 = 6(-2) – 9
- 21 = - 12 – 9
- 21 = - 21
L.H.S = R.H.S ⇒ yes a solution
<em><u>Option D)</u></em>
- 21 = 6(9) – 9
- 21 = 54 – 9
- 21 = 45
L.H.S ≠ R.H.S ⇒ not a solution
Hence, the solution for the given equation is – 2, so option c is correct
Answer:
14?
Step-by-step explanation:
Answer:
You can split your cracker 4 ways. So each cracker is now 4. Now split your cracker into 4ths, 5•4 = 20! Hope this helps!
Step-by-step explanation:
Your Crackers have those lines, and they split 4 ways. SO this helps make the problem easierto solve.
