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olasank [31]
3 years ago
10

10n + 7 > n - 3 (5 - 3n) plz

Mathematics
2 answers:
ratelena [41]3 years ago
5 0
So whats the question?
Ad libitum [116K]3 years ago
4 0
Are you trying to figure out the value of n?
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Mr. Gonzales has only $42.50 to spend at a clothing store. He wants to buy a shirt that costs $29, including tax, and some brace
lisabon 2012 [21]

Answer: The equation is $29 + x*$4.50 = $42.50, and the solution is x = 3

Step-by-step explanation:

The data we have is:

Gonzales has $42.50

He wants to buy:

a shirt that costs $29

some bracelets that cost $4.50 each.

The equation that we need to solve is:

Total cost = money that Gonzales has.

The cost is $29 + x*$4.50

where x is the number of bracelets he can buy.

The equation that we need to solve is:

$29 + x*$4.50 = $42.50

to solve it we must isolate x:

x*$4.50 = $42.50 - $29 = $13.50

x = 13.50/4.50 = 3

So we have that Mr. Gonzales can buy a total of 3 bracelets.

8 0
3 years ago
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Please help me out with number 7
Slav-nsk [51]
The answer for number 7 is x=3

4 0
3 years ago
Which number is prime 15, 24, 41, 62
Juli2301 [7.4K]
Forty-one is a prime number because it can multiply itself by one.
7 0
3 years ago
WHOEVER gives the CORRECT answer FIRST, gets brainliest answer and thanks
Alex_Xolod [135]
Yes, because they are similar(or you can call congruent), proved by SAS or simply HL.
7 0
3 years ago
Read 2 more answers
Remember to show work and explain. Use the math font.
vichka [17]

Answer:

\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}

Step-by-step explanation:

(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}

2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------

3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}

7 0
3 years ago
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