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n200080 [17]
3 years ago
13

Need help with this solve for x question!

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
7 0

Answer:

x=4

Step-by-step explanation:

RS + ST = RT

5+ 2x-2 = 3x-1

Combine like terms

2x+3 = 3x-1

Subtract 2x from each side

2x+3-2x = 3x-1-2x

3 =x-1

Add 1 to each side

3+1 = x-1+1

4=x

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3 - 2n=1 - 2n<br> Lo nesito
omeli [17]

Answer:

0=−2

Step-by-step explanation:

espero que esto ayude! y que Dios te bendiga .. mantente a salvo!

hope this is helpful God Bless you stay safe!

6 0
3 years ago
Read 2 more answers
Giventheequation3​x​+​y​=6,writeasecondequationthat,togetherwiththefirst,will create a system of equations thata.has one solutio
Svetradugi [14.3K]
(4,7.8) is your answer please give brainliest!
6 0
3 years ago
A cake decorator rolls a piece of stiff paper to form a cone. She cuts off the tip of the cone and uses it as a funnel to pour d
Nimfa-mama [501]

Answer:

Step-by-step explanation:

Use the formula for volume of a cone

V = 1/3 pi r^2 (h)

V = 1/3 pi 66^2 (18)

V= 1/3 pi 4356(18)

V = 1/3 pi 78408

V = 1/3 264,201.12

V = 82,067.04cm^3

5 0
2 years ago
#12 with explanation please
4vir4ik [10]

Answer:

  • 30.01 units

Step-by-step explanation:

<u>Given vertices WXYZ:</u>

  • W(-4,5), X(2,4), Y(3, -4), Z(-1, -6)

Perimeter is the sum of the side length.

<u>Use distance formula to find each side:</u>

  • WX = \sqrt{(2+4)^2+(4-5)^2} =\sqrt{37} ≈ 6.08
  • XY = \sqrt{(3-2)^2+(-4-4)^2} =\sqrt{65} ≈ 8.06
  • YZ = \sqrt{(-1-3)^2+(-6+4)^2} =\sqrt{20} ≈ 4.47
  • WZ = \sqrt{(-1+4)^2+(-6-5)^2} =\sqrt{130}\sqrt{(-1+4)^2+(-6-5)^2}=\sqrt{130} ≈ 11.40

<u>Find perimeter:</u>

  • P = 6.08 + 8.06 + 4.47 + 11.40 = 30.01 units
7 0
3 years ago
What are the potential solutions to the equation 2ln(x+3)=0
skad [1K]
2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1

Expanding the left side gives

x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0

which gives two solutions, x=-4 and x=-2. But if x=-4, then \ln(x+3)=\ln(-1), but this number isn't real, so x=-4 is an extraneous solution. Meanwhile if x=-2, you get \ln(-2+3)=\ln1=0, so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.
4 0
3 years ago
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