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olga55 [171]
3 years ago
15

Help me please I really need help

Mathematics
2 answers:
Crank3 years ago
8 0
I think the answer is A
lianna [129]3 years ago
7 0

Answer:

im pretty sure its A

Step-by-step explanation:

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Hi, can someone please give me a detailed explanation on how to solve this problem?
Dovator [93]

Answer:

Step-by-step explanation:

log3^(2x+3) = log243

(2x+3)log3 = log243

2x+3 = log243/log3

2x = (log243/log3)-3

x = [(log243/log3)-3]/2

x = (5-3)/2

x = 2/2

x = 1

Check:

3^2(1)+3 = 243

3^2+3 = 243

3^5 = 243

3×3×3×3×3 = 243

9×9×3 = 243

81×3 = 243

243 = 243

5 0
3 years ago
At the pie eating contest, Max ate 2⁄5 of a pie and Jen ate 6⁄8 of a pie. How many pies did they eat all together?
Brums [2.3K]
I think you have to find a common multiple for the denominators 5 and 8 which would be 40. so it would be 6/40 and 2/40. then I think, you  would add or multiple, depending on the situation, 6 and 2
6 0
3 years ago
Read 2 more answers
Please answer this question above.
ioda
The distance between point J and line KL would be the line perpendicular to the line, which is line JM.

The answer would be B.
7 0
3 years ago
Two possible solutions of √11-2x=√x^2+4x+4 are –7 and 1. Which statement is true? A) Only x = –7 is an extraneous solution.
11111nata11111 [884]

Answer:

Option D)Neither solution is extraneous.

Step-by-step explanation:

we have

\sqrt{11-2x}=\sqrt{x^{2}+4x+4}

we know that

two possible solutions are x=-7 and x=1

<u><em>Verify each solution</em></u>

Substitute each value of x in the expression above and interpret the results

1) For x=-7

\sqrt{11-2(-7)}=\sqrt{-7^{2}+4(-7)+4}

\sqrt{25}=\sqrt{25}

5=5 ----> is true

therefore

x=-7 is not a an extraneous solution

2) For x=1

\sqrt{11-2(1)}=\sqrt{1^{2}+4(1)+4}

\sqrt{9}=\sqrt{9}

3=3 ----> is true

therefore

x=1 is not a an extraneous solution

therefore

Neither solution is extraneous

7 0
3 years ago
Read 2 more answers
Select the correct answer.
Paul [167]
The answer should be. 5x=-12
3 0
3 years ago
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