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Agata [3.3K]
2 years ago
13

Zoe placed colored blocks on a scale in science class. Each block weighed 0.8 ounces. The total weight of all the colored blocks

was 12.8 ounces. How many blocks did Zoe place on the scale? Write and solve an equation to find the answer. number of blocks =
Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
3 0

Answer:

16 blocks. 12.8=(.8)x

Step-by-step explanation:

ale4655 [162]2 years ago
3 0

Answer:

16

Step-by-step explanation:

because it both so it come out to be 16

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Marina86 [1]

The answer is 52,000.


You have to add all of the values and divide them by the number of values. In this case you would do 30,000 + 50,000 + 30,000 + 80,000 + 70,000 which equals 260,000 then you would divide that number by 5 since there are 5 values.

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3 years ago
What is the value of y in the equation 2(3y - 6) = 0?
Fittoniya [83]

Answer:

The answer is y=2

Step-by-step explanation:

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Can someone help me with this i have been trying for hours and i actually got the wrong answer so can you help me understand by
kap26 [50]

Answer:

C

Step-by-step explanation:

We are given that FGH is a triangle, which means that...

Angle F + Angle G + Angle H = 180

Start by substituting the given angle measurements into the equation:

4x+2+13x-7+3x+5=180

Combine like terms

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Divide both sides by 20 to isolate x

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Plug 9 back in for x to solve for each angle

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Angle G = 13(9)-7=117-7=110

Angle H = 3(9)+5=27+5=32

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Step-by-step explanation:

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1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
                                                     0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
                             x = 0
5 0
3 years ago
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