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NeX [460]
3 years ago
12

120x=120. X= PLS HELP ME IM IN A DEADLINE AND DUNNO HOW TO DO THIS

Mathematics
1 answer:
Butoxors [25]3 years ago
6 0
X=1
jsjeiaiajahahajajja
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Andy estimated that he would need 79 feet of lumber for a tree house project. He later found that the actual amount of lumber ne
Radda [10]

Answer:

Step-by-step explanation:

79 divided by 68 is your answer

You can do the math

5 0
3 years ago
Need help with geometry question?
Nuetrik [128]

Answer: A, Corresponding; Vertical


Step-by-step explanation: The answer for angles 1 and 5 are Corresponding Angles (Elevator Angles)  and angles 1 and 4 are Vertical Angles.

All Angles besides a Linear Pair and Same-Side Interior Angle are Congruent.

5 0
3 years ago
Ms. Walker's class set up an online fund with a goal to raise $1,280 to go on a field trip. Ms. Walker starts the fund by deposi
madreJ [45]
What you can use for this case is a function of the potential type.
 We have then
 y = a (b) ^ x
 Where we have:
 Walker starts the fund by depositing $ 5
 a = 5
 Each week the balance of the fund is twice the balance of the previous week:
 b = 2
 The function is:
 y = 5 (2) ^ x
 The number of weeks to reach $ 1280 is 8 weeks.
 Check:
 y = 5 (2) ^ 8
 y = 1280
 Answer:
 An equation can be used to find the number of weeks, x, after which the balance of the fund will reach $ 1,280 is:
 y = 5 (2) ^ x
 The number of weeks that it takes to reach the class goal is
 8 weeks
4 0
3 years ago
Read 2 more answers
Venetta buys 2 pounds of pistachios and 3 pounds of almonds. The pistachios cost $4 more per pound than the almonds. She pays a
Alex

Answer: the answer is 2 and 5

Step-by-step explanation:

8 0
3 years ago
Determine the zeroes of the polynomial
Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)}  \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)}  -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\

\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
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