9 batches because if you kept counting until you got close to 12 cups of flour you would get 9 batches
1. solve for x in first to just find nterms of y
x=12-y
sub that for x in other one
first divide that one by 2 to make it easier
2x-y=18
2(12-y)-y=18
24-2y-y=18
24-3y=18
minus 24 both sides
-3y=-6
divide -3
y=2
D is answer
2.
find intersection of x+y=5 and -2x+3y=6
or easiers, just test the intersections
wait, they all intersect in the same place
this is super easy
pick ANY point that is in that reigon and see if it is true
A, if we pick lets say (5,5)
that is false for x+y<5
not A
B
pick (10,0)
false for first one
not B
C
pick (0,0)
true for first
false for 2nd
not C
D. pick (-10,0)
true for first
true for 2nd
answer is D
answer is D for both questions
Solution is x = -2.2
<h2>Linear system</h2>
It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.
<h3>Simplification</h3>
Simplification is to make something easier to do or understand and to make something less complicated.
Given
The linear equation 0.45x + 0.33 = -0.66
<h3>To find </h3>
The value of x.
<h3>How to find the value of x?</h3>
The linear equation 0.45x + 0.33 = -0.66 is given.
On simplifying,
0.45x + 0.33 = -0.66
0.45x = -0.66 - 0.33
0.45x = -0.99
x = -2.2
Hence, the value of x is -2.2.
Solution is x = -2.2
More about the linear system link is given below.
brainly.com/question/20379472
Answer:
693 Years
Step-by-step explanation:
Given an initial amount
and k (a negative constant) determined by the nature of the material, the amount of radioactive material remaining at a given time t, is determined using he formula:
![A(t)= A_oe^{kt}](https://tex.z-dn.net/?f=A%28t%29%3D%20A_oe%5E%7Bkt%7D)
If a certain radioactive isotope decays at a rate of 0.1% annually.
![\frac{1}{2} A_o= A_oe^{-0.001t}\\e^{-0.001t}=\frac{1}{2}\\$Take the natural logarithm of both sides\\-0.001t=ln(0.5)\\t=ln(0.5) \div -0.001\\t=693.15\approx 693 \:years](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20A_o%3D%20A_oe%5E%7B-0.001t%7D%5C%5Ce%5E%7B-0.001t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C%24Take%20the%20natural%20logarithm%20of%20both%20sides%5C%5C-0.001t%3Dln%280.5%29%5C%5Ct%3Dln%280.5%29%20%5Cdiv%20-0.001%5C%5Ct%3D693.15%5Capprox%20693%20%5C%3Ayears)
The half-life of the isotope is 693 years.