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shusha [124]
3 years ago
5

2.7 (7.8)??? I don't know how to work this problem...

Mathematics
2 answers:
ruslelena [56]3 years ago
8 0
Just multiply the two, your answer should be 21.06
12345 [234]3 years ago
7 0
The answer is 21.06 easy right
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Find two numbers with a sum of 20 and a difference of 14.
vivado [14]
A and B Definitely. 3 and 17, -3 and -17
4 0
2 years ago
Andrew answered 21 out of 25 questions correctly on his math test. Written as a decimal, what portion of Andrew's answers were c
soldier1979 [14.2K]
Since we are calculating the amount that he got correct, we must divide the numerator by the denominator or make the denominator equal 100. I will do it the second way as it is easier in this case.

25*4 = 100
21*4 = 84

84/100 = 0.84

As a decimal, Andrew got 0.84 of his test correct. Hope this helps!
8 0
3 years ago
The perimeter of ABCD is 66 and DC is twice as long as CB how long is a B
Natali5045456 [20]
<span>Length = x Width = 2x
Perimeter = 2 * length + 2 * width
= 2x + 4x
= 6x 
6x = 66
x = 11 
</span><span>Length = x Width = 2x
Perimeter = 2 * length + 2 * width
= 2x + 4x
= 6x 
6x = 66
x = 11</span>
3 0
3 years ago
Help me!!!!!!!!!!!!!
Stells [14]

Answer:

Both of the lines that goes to A is the same length so just find the length measurement of the line fro C to A and the write that as the answer.

Step-by-step explanation:


7 0
3 years ago
The hydrogen ion concentration (h+) in a certain cleaning compound is (h+)=3.2x10^-11 use the formula ph=-log(h+) to find the pH
Furkat [3]

Answer:

The pH of the clean compound is approximately 10.495.

Step-by-step explanation:

If the cleaning ion concentration of the cleaning compound is [H^{+}] = 3.2\times 10^{-11}, then the pH, no unit, of the compound is:

pH = -\log_{10} [H^{+}]  (1)

pH = -\log_{10} (3.2\times 10^{-11})

pH = -\log_{10} 3.2 - \log_{10} (10^{-11})

pH = -\log_{10} \frac{32}{10} +11

pH = -\log_{10} 32 + \log_{10} 10 + 11

pH = -\log_{10} 32 +12

pH \approx 10.495

The pH of the clean compound is approximately 10.495.

6 0
2 years ago
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