Please help, due today

Mathematics

1 answer:

kaheart [24]3 years ago

4 0

Answer:

See below.

Step-by-step explanation:

The domain refers to the x values used in the equation/graph. The line appears to go from -10 to 10 on the x-axis. The dots above -10 and 10 are not shaded in meaning those x values do not count towards the domain meaning we will use parenthesis instead of brackets for interval notation or no equal to signs in inequality notation.

Interval Notation: (-10, 10)

Inequality Notation: -10 < x < 10

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a spherical balloon is deflated at a rate of 256pi/3 cm^3/sec. at what rate is the radius of the balloon changing when the radiu

Hunter-Best [27]

What you need to look for is $\frac { dr }{ dt }$ when r=8.

Now:

$Volume\quad of\quad a\quad sphere:\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }$

$\therefore \quad \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi \cdot 3{ r }^{ 2 }=4\pi { r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dV } =\frac { 1 }{ \frac { dV }{ dr } } =\frac { 1 }{ 4\pi { r }^{ 2 } }$

$And:\\ \\ \frac { dV }{ dt } =-\frac { 256\pi }{ 3 } \\ \\ Therefore:\\ \\ \frac { dr }{ dt } =\frac { dr }{ dV } \cdot \frac { dV }{ dt }$

$\\ \\ =\frac { 1 }{ 4\pi { r }^{ 2 } } \cdot -\frac { 256\pi }{ 3 } \\ \\ =-\frac { 256 }{ 12 } \cdot \frac { \pi }{ \pi } \cdot \frac { 1 }{ { r }^{ 2 } }$

$\\ \\ =-\frac { 64 }{ 3{ r }^{ 2 } } \\ \\ When\quad r=8,\\ \\ \frac { dr }{ dt } =-\frac { 64 }{ 3\cdot { 8 }^{ 2 } } =-\frac { 1 }{ 3 } \\ \\ Answer:\quad -\frac { 1 }{ 3 } \quad cm/sec$

3 0

4 years ago

Hey so im also looking for the first answer and the choices were “ is not “ and “ is “

dexar [7]

so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}$

nope, the analysis shows that Charlie was going faster than 35 m/h.

now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d$