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3 years ago

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 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

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Giventheequation3x+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!

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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:

Given vertices WXYZ:

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

Perimeter is the sum of the side length.

Use distance formula to find each side:

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

Find perimeter:

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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