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EastWind [94]
3 years ago
14

One day each summer, Chandler earns money by having a lemonade stand in his neighborhood. He charges $0.50 per glass of lemonade

, and he made a total of $24. How many glasses of lemonade did he sell?
Mathematics
2 answers:
alex41 [277]3 years ago
5 0
He sold 48 glasses of lemonade.
VLD [36.1K]3 years ago
4 0
The equation would be .50x=24

We would divide both sides by .50 so that way we have x by itself

.50x/.50=x
24/.50=48
So x=48
So he sold 48 glasses of lemonade
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Inessa05 [86]
The answer to that is -40
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When -2(x − 4) + 1 = 7 is solved, the result is:
Sergeu [11.5K]

Answer:

1

Step-by-step explanation:

-2(x-4)+1=7

First Distribute the -2.

-2x+8+1=7

Subtract the 8 and 1 from the whole equation.

-2x=-2

Divide both sides of the equation by -2.

x=1

I hope this helps!

6 0
3 years ago
Read 2 more answers
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

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

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

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

6 0
2 years ago
Cx=r/d<img src="https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7Bd%7D" id="TexFormula1" title="\frac{r}{d}" alt="\frac{r}{d}" align="abs
antoniya [11.8K]

Answer:

x=\frac{rc}{d}

Step-by-step explanation:

ill take the equation as this

cx=\frac{r}{d}

divide both side by c

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}

Next time when you see someone greif your question, please report it.

Shame for the other guy

Hoped this helped you

Red

8 0
3 years ago
HELPPPPPPPPPPPPPPPPPP
alexira [117]
It’s the 2nd option :)
7 0
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