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Bumek [7]
4 years ago
6

P - 198,23 = 127.69A. p = 316.92B. p = 325.92C. p = 325.82

Mathematics
1 answer:
tia_tia [17]4 years ago
8 0
The answer in most likely B
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What values of a and b would make the equation shown have infinitely many solutions? 3x=ax+b
Wittaler [7]
Infinitely many solutions means that you have the same thing on both sides of the equation no matter what value of x you plug in, right?

We just need both sides to be 3x then, correct?

If a were equal to 3 and b were equal to 0, we'd have

3x = (3)x + 0

Which is essentially 3x = 3x

So that means a = 3 and b = 0 must work!
Let's say x = 5

3(5) = 3(5) + 0

15 = 15 + 0

15 = 15

That means that a = 3 and b = 0 is your final answer :)
7 0
3 years ago
the soccer field at Mario's school has an area of 6,000 square meters. how can mario show the area as a whole number multiplied
ruslelena [56]

6,000 = 6 x 10^3 square meters.
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3 0
3 years ago
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Help me out please! Any unrelated answers will be reported, so don't try :)
timama [110]

Try doing A, 2/3 Hopefully it helps you, I'm not good with fractions.

8 0
4 years ago
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Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)
Tems11 [23]

The slope of the tangent line to f at x=-1 is given by the derivative of f at that point:

f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}

Factorize the numerator:

2x^2-2=2(x^2-1)=2(x-1)(x+1)

We have x approaching -1; in particular, this means x\neq-1, so that

\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)

Then

f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4

and the tangent line's equation is

y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2

6 0
3 years ago
Plzzz Select the correct answer.
krek1111 [17]

Answer: D:

To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 5. The solution to system B will be the same as the solution to system A.

Step-by-step explanation:

7 0
3 years ago
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