The pounds of alloy that contains 26% copper that would be used is 23.42 pounds.
The pounds of alloy that contains 69% copper that would be used is 29.58 pounds.
<h3>What are the linear equations that represent the question</h3>
0.26a + 0.69b = (53 x 0.5)
0.26a + 0.69b = 26.50 equation 1
a + b = 53 equation 2
Where:
- a =pounds of alloy that contains 26% copper
- b = pounds of alloy that contains 69% copper
<h3>How many pounds of each alloy should be in the third alloy?</h3>
Multiply equation 2 by 0.26
0.26a + 0.26b = 13.78 equation 3
Subtract equation 3 from equation 2
12.72 = 0.43b
b = 12.72 / 0.43
b = 29.58 pounds
a = 53 - 29.58 = 23.42 pounds
To learn more about linear functions, please check: brainly.com/question/26434260
#SPJ1
Answer:
BC = q +p
Step-by-step explanation:
<h3>b)</h3>
Here's one way to get there. Two vector sums are the same if they have the same starting and ending points. The sum AC can be expressed two ways:
AB +BC = AD +DC
p +BC = q +2p . . . . . . using givens, and your result from part (a)
Subtract p from both sides:
BC = q +p
Answer:
1. False 2. 7.11 > -7.1 3. -3 < -1 and -3 <= -1 <= is less than or equal to. 4. m + 7 => 8.2 5. {3, 4} Hope this helps even though I am late!
Step-by-step explanation:
F(x) = -2 (x-9)2 + 14
= -4(x-9) + 14
= -2 [ 2(x-9) -7 ]
= -2 (2x -18 -7)
= -2 (2x -25)
