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3 years ago

14

bill spent $19 at starbucks on a coffee and some cookies. if the coffee cost $5 each cookie cost $2, how many cookies did he buy

?

1 answer:

Nataliya [291]3 years ago

6 0

Answer:

this is easy. we know that he spent $19 total, 1 coffee that costed $5, and that each cookie costs $2. all you have to do is subtract 5 from 19 and divide the remaining amount of money (14) by 2. this makes 7 the amount of cookies Bill bought.

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Which values of c will cause the quadratic equation -- x2 + 3x + c = 0 to have no real number solutions? Check all that

Novay_Z [31]

Answer:

c=5x

Step-by-step explanation:

:D good luck

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=log_%7B8%20x%5E%7B2%7D%20-23x%2B15%7D%282x-2%29%20%5Cleq%200" id="TexFormula1" title="log_{8 x

grandymaker [24]

$\log_{8x^2-23x+15} (2x-2) \leq 0$

The domain:
The number of which the logarithm is taken must be greater than 0.
$2x-2 \ \textgreater \ 0 \\
2x\ \textgreater \ 2 \\
x\ \textgreater \ 1 \\ x \in (1, +\infty)$

The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:
$8x^2-23x+15\ \textgreater \ 0 \\ 8x^2-8x-15x+15\ \textgreater \ 0 \\ 8x(x-1)-15(x-1)\ \textgreater \ 0 \\ (8x-15)(x-1)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ 8x-15=0 \ \lor \ x-1=0 \\ 8x=15 \ \lor \ x=1 \\ x=\frac{15}{8} \\ x=1 \frac{7}{8} \\ \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola op} \hbox{ens upwards} \\
\hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros} \\ \\
x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty)$

*not equal to 1:
$8x^2-23x+15 \not= 1 \\
8x^2-23x+14 \not= 0 \\
8x^2-16x-7x+14 \not= 0 \\
8x(x-2)-7(x-2) \not= 0 \\
(8x-7)(x-2) \not= 0 \\
8x-7 \not=0 \ \land \ x-2 \not= 0 \\
8x \not= 7 \ \land \ x \not= 2 \\
x \not= \frac{7}{8} \\ x \notin \{\frac{7}{8}, 2 \}$

Sum up all the domain restrictions:
$x \in (1, +\infty) \ \land \ x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty) \ \land \ x \notin \{ \frac{7}{8}, 2 \} \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2) \cup (2, +\infty)
$

The solution:
$\log_{8x^2-23x+15} (2x-2) \leq 0 \\ \\
\overline{\hbox{convert 0 to the logarithm to base } 8x^2-23x+15} \\
\Downarrow \\
\underline{(8x^2-23x+15)^0=1 \hbox{ so } 0=\log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ }
\\ \\
\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1$

Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.

* if the base is less than 1:
$8x^2-23x+15 \ \textless \ 1 \\
8x^2-23x+14 \ \textless \ 0 \\ \\
\hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\
\hbox{the values less than 0 are between the zeros} \\ \\
x \in (\frac{7}{8}, 2) \\ \\
\hbox{including the domain:} \\
x \in (\frac{7}{8}, 2) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (1 \frac{7}{8} , 2)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{flip the sign} \\ 2x-2 \geq 1 \\ 2x \geq 3 \\ x \geq \frac{3}{2} \\ x \geq 1 \frac{1}{2} \\ x \in [1 \frac{1}{2}, +\infty) \\ \\ \hbox{including the condition that the base is less than 1:} \\ x \in [1 \frac{1}{2}, +\infty) \ \land \x \in (1 \frac{7}{8} , 2) \\ \Downarrow \\ x \in (1 \frac{7}{8}, 2)$

* if the base is greater than 1:
$8x^2-23x+15 \ \textgreater \ 1 \\ 8x^2-23x+14 \ \textgreater \ 0 \\ \\ \hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\ \hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros}$

$x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \\ \\ \hbox{including the domain:} \\ x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (2, \infty)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \\ x \leq 1 \frac{1}{2} \\ x \in (-\infty, 1 \frac{1}{2}] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 \frac{1}{2}] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset$

Sum up both solutions:
$x \in (1 \frac{7}{8}, 2) \ \lor \ x \in \emptyset \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2)$

The final answer is:
$\boxed{x \in (1 \frac{7}{8}, 2)}$

5 0

3 years ago

Can i have help with this

lilavasa [31]

Answer:

Option D

Step-by-step explanation:

<u>Square:</u>

Side = 25 ft

Area of square = side * side

= 25*25

= 625 square ft

<u>Area of semicircle:</u>

diameter = 25 ft

r = 25/2 = 12.5 ft

$\boxed{\text{Area of semicircle =$ \dfrac{\pi r^{2}}{2} $}}$

$=\dfrac{3.14 * 12.5*12.5}{2}\\\\\\= 245.3 \ square \ ft$

Area of the shaded region = area of square - area of semicircle

= 625 - 245.3

= 379.7 square ft

8 0

3 years ago

G(x)=1/6(8)^x-5. x=9

Nezavi [6.7K]

Answer:

$g(9) = \frac{2048}{3}$

Step-by-step explanation:

Given

$g(x) = \frac{1}{6}(8)^{x-5}$

$x = 9$

Required

Solve

Substitute 9 for x in $g(x) = \frac{1}{6}(8)^{x-5}$

$g(9) = \frac{1}{6}(8)^{9-5}$

$g(9) = \frac{1}{6}(8)^{4}$

Remove bracket

$g(9) = \frac{1}{6}* 8^4$

$g(9) = \frac{1}{6}* 4096$

$g(9) = \frac{1* 4096}{6}$

$g(9) = \frac{4096}{6}$

Simplify fraction

$g(9) = \frac{2048}{3}$

5 0

3 years ago

Franco hikes 4 miles each day for 5 days. He carries 100 ounces of water with him. It takes him 1hour to hike 4miles. How many h

liberstina [14]

20 hours because 1 mile is 1 hour so 4 miles is 4 hours so if you times the 5 days by the 4 miles/ hours you get 20 hours in 5 days.

6 0

3 years ago