I need help with this math problem
1 answer:
Consider the right triangle HBF. The Pythagorean theorem tells you ...
HF² = HB² + BF²
The lengths HB and BF can be determined by counting grid squares, or by subtracting coordinates. Here, it is fairly convenient to count grid squares. When we do that, we find ...
HB = 2
BF = 5
Using these values in the equation above, we get
HF² = 2² + 5²
HF² = 4 + 25 = 29
Taking the square root gives the length HF.
HF = √29
HF² = HB² + BF²
The lengths HB and BF can be determined by counting grid squares, or by subtracting coordinates. Here, it is fairly convenient to count grid squares. When we do that, we find ...
HB = 2
BF = 5
Using these values in the equation above, we get
HF² = 2² + 5²
HF² = 4 + 25 = 29
Taking the square root gives the length HF.
HF = √29
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Answer:
<h2>3x - 2</h2>Step-by-step explanation:
<em>The distributive property</em>:
<em>a(b + c) = ab + ac</em>
<h3>Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
_____
<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy