The area of ΔIDK is 32.69 square units.
Solution:
Given data:
ID = 7 and Hypotenuse, KI = 11.67.
Let us first find KD:
Using Pythagoras theorem,
Subtract 49 from both sides.
Taking square root on both sides, we get
KD = 9.34
Area of the triangle = 
= 32.69
The area of ΔIDK is 32.69 square units.
It should be the second answer.
Reason:
-np-3≥7(c-4)
-np≥7c-28+3
-np≥7c-25
-n≥(7c-25)/p
N≤-(7c-25)/p
"At least" refers to greater than or equal to (≥).
This is because "at least" means:
- it can't be less than
- it can be equal to
- it can be greater than
This is basically like saying no less than.
If you must have at least 10 oranges for some recipe, then you must have 10 oranges, but you can also have more.
"Twice the difference of a number and 8"
This means that 2 times a number minus 8.
This can be written as 2(w - 8), since w is unknown and it's the difference between w and 8.
The 2 is being multiplied by this because it's 2 times this difference.
"At least 26"
This means ≥ 26.
Now for the actual inequality.
"Twice the difference of a number and 8"
translates to 2(w - 8)
"is at least 26"
translates to ≥ 26
Now put them together.
2(w - 8) ≥ 26
So the answer is 2(w - 8) ≥ 26.
Hope this helps! :)
Help with what? there’s nothing?
Steps:
1. Take -x + 2y= 21
2. Add x on both sides
2y= x + 21
3. Divide everything by 2
Y= 1/2x + 10.5
4. Substitute the y in 7x + 2y= -19 for 1/2x + 10.5
7x + 2(1/2x + 10.5)= -19
5. Multiply everything in parentheses
7x + x + 21= -19
8x + 21= -19
6. Subtract 21 on both sides
7. 8x=-40
8. Divide each side be 8
9. Your answer is
X=-5
