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3 years ago

15x>-60 help me please

Mathematics

1 answer:

devlian [24]3 years ago

5 0

Answer:

x>4

Step-by-step explanation:

15x>-60

x>-60/15

x>4

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How many batches of cookies can she make if she had 7 1/2 cup of sugar

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3 years ago

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Solve for x. -(-2-5x) + (-2) = 18 x= -90 x=-18/5 2 = 18/5 x= 90

mart [117]

Answer:

$x=\frac{18}{5}$

Step-by-step explanation:

$-\left(-2-5x\right)+\left(-2\right)=18$

Add 2 to both sides:

$-\left(-2-5x\right)-2+2=18+2$

$-\left(-2-5x\right)=20$

Divide both sides by -1:

$\frac{-\left(-2-5x\right)}{-1}=\frac{20}{-1}$

$-2-5x=-20$

Add 2 to both sides:

$-2-5x+2=-20+2$

$-5x=-18$

Divide both sides by -5:

$\frac{-5x}{-5}=\frac{-18}{-5}$

$x=\frac{18}{5}$

___________________________

OAmalOHopeO

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3 years ago

End behavior of polynomials I give 20 points

raketka [301]

The end behavior of a polynomial is determined entirely by the leading term. For example, $2x^4-x+1$ behaves the same way as $2x^4$ when $x$ is very large in magnitude. All the problems in the picture are more or less the same. I'll pick out two examples:

1a) $f(x)=5x^6-3x^3-4x^2+8x$ behaves like $5x^6$ . $x^6$ will be positive regardless of the value of $x$ , so to either the left or right, the graph of $f(x)$ will approach positive infinity or "rise" on both the left and right sides.

1c) $f(x)=-4x(x-4)(x+2)$ has a leading term of $-4x^3$ , so $f(x)$ behaves like $-4x^3$ . When $x$ , we have $x^3$ , so that $-4x^3>0$ . This means $f(x)$ rises to the left. If $x>0$ , then $-4x^3$ , so $f(x)$ falls to the right.