Answer:
Step-by-step explanation:
1). ∵ 1 L = 1000 ml
∴ 4 L = 4000 ml
2). ∵ 1 cm = 10 mm
∴ 5 cm = 50 mm
3). ∵ 1 mg = 0.1 cg
∴ 5 mg = 0.5 cg
4). ∵ 1 ml = 0.001 L
∴ 25 ml = 0.025 L
5). ∵ 1 gm = 1000 mg
∴ 0.125 gm = 25 mg
6). ∵ 1 ml = 0.001 L
∴ 570 ml = 0.57 L
7). ∵ 1 mm = 0.1 cm
∴ 135 mm = 13.5 cm
8). ∵ 1 mg = 0.001 g
∴ 1820 mg = 1.82 g
9). ∵ 1 g = 100 cg
∴ 0.5 g = 50 cg
10). ∵ 1 ml = 0.001 L
∴ 0.2385 ml = 0.0002385 L
18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.
<u>Step-by-step explanation:</u>
Let x = kg of 15% copper alloy
Let y = kg of 60% copper alloy
Since we need to create 90 kg of alloy we know:
x + y = 90
51% of 90 kg = 45.9 kg of copper
So we're interested in creating 45.9 kg of copper
We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:
0.15x + 0.60y = 45.9
but
x + y = 90
x= 90 - y
substituting that value in for x
0.15(90 - y) + 0.60y = 45.9
13.5 - 0.15y + 0.60y = 45.9
0.45y = 32.4
y = 72
Substituting this y value to solve for x gives:
x + y = 90
x= 90-72
x=18
Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.
Answer:
x=−1 or x=−9
Step-by-step explanation:
x2+11x+10=x+1
Step 1: Subtract x+1 from both sides.
x2+11x+10−(x+1)=x+1−(x+1)
x2+10x+9=0
Step 2: Factor left side of equation.
(x+1)(x+9)=0
Step 3: Set factors equal to 0.
x+1=0 or x+9=0
x=−1 or x=−9
Answer:
29
Step-by-step explanation:
0.11v=3.19

Divide:
3.19÷0.11=29
v=29
__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)____φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)v__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)__φ(。。)