2 years ago

Alexis has 6 more collector's cards than Danny. Together, the two friends have 24

Mathematics

2 answers:

Pavel [41]2 years ago

7 0

Answer:

Step-by-step explanation:

notsponge [240]2 years ago

5 0

Answer:

Step-by-step explanation:

24-6

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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.

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3 years ago

Read 2 more answers

2n-n-4+7n in simplest form

MrMuchimi

<span>2n-n-4+7n in simplest form is found by combining like terms:

(2-1+7)n - 4 = 8n-4, or 4(2n-1)</span>

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3 years ago

Are the triangles congruent if they are which theorem and why?

leonid [27]

Yes the sides are equal to each other

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