2x - 4y + 1z = 11 ⇒ 2x - 4y + 1z = 11
1x + 2y + 3z = 9 ⇒ <u>1x + 2y + 3z = 9</u>
3x + 5z = 12 1x - 2y - 2z = 2
1x - 2y - 2z = 2
2x - 4y + 1z = 11 <u>-2x + 2y - 2z = -3</u>
1x + 2y + 3z = 9 ⇒ 1x + 2y + 3z = 9 x - 4z = 1
3x + 5z = 12 ⇒ <u>3x + 5z = 12</u> x - 4z + 4z = 1 + 4z
-2x + 2y - 2z = -3 x = 1 + 4z
<u /> 1 + 4z - 2y - 2z = 2<u />
1 - 2y + 4z - 2z = 2
1 - 2y + 2z = 2
<u>- 1 - 1</u>
-2y + 2z = 1
-2y + 2z - 2z = 1 - 2z
<u>-2y</u> = <u>1 - 2z</u>
-2 -2
y = -0.5 + z
x + 2(-0.5 + z) - 2z = 2
x - 1 + z - 2z + 2 = 2
x - 1 + z = 2
<u> + 1 + 1</u>
x + z = 3
x - x + z = 3 - x
z = 3 - x
<u />
Answer:
Part A; Initial dosage is 10 milligrams
Part B:
8.4113 milligrams after 1 hour
5.0057 milligrams after 4 hours.
Step-by-step explanation:
It is given M(h)= 10
Initial dosage is the 10 milligrams
After 1 hour, plug in h as 1
So, M(1)=10
Simplify the right side
M(1)=10(0.84113)
M(1)= 8.4113 milligrams.
Now, after 4 hours
M(4)=10
M(4)=10(0.50057)
M(4) =5.0057 milligrams
Answer:
Step-by-step explanation:
Given:
Consider the completer question is "If ∆BTS≅∆GHD, BS=25, TS=14, BT=31, GD=4x-11, m∠S=56, m∠B=21 and m∠H=(7y+5), find the values of x and y.
To find:
The values of x and y.
Solution:
We have,
(Given)
(CPCTC)
Divide both sides by 4.
In ∆BTS,
(Angle sum property)
Now,
(CPCTC)
Divide both sides by 7.
Therefore, the value of x is 9 and value of y is 14.