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2 years ago

Solve the inequality. Graph the solution. x +19]>4

Mathematics

1 answer:

Yakvenalex [24]2 years ago

4 0

Answer:

x>-15

Step-by-step explanation:

if:x+19>4

collect like terms

therefore:

x>4-19

(sign changes when collecting like terms)

+4-19=-15

x>-15

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Lines A, B, and C show proportional relationships. Which line has a constant of proportionality between y and x of 4/3

Drupady [299]

Answer:

Line B

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3 years ago

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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:

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

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

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

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

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

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3 years ago

Find the equation of the linear function 'g' going through (20,5) and (-5,-8). Use 'u' for your input variable.

frosja888 [35]

Answer:

See below

Step-by-step explanation:

Use the slope-intercept form, which will be in this case:

g(u) = mu + b, where m is the slope, b is the y- intercept

Find the slope:

$m=(y_2-y_1)/(x_2-x_1)$
$m=(-8-5)/(-5-20)=(-13)/(-25)=13/25$

Find the value of the y-intercept:

$5=1/25*20+b$
$b=5-4/5=21/5$

The equation is:

$g(u)=\frac{13}{25} u+\frac{21}{5}$

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2 years ago

Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different

N76 [4]

Answer:

(a) As time increases, the amount of water in the pool increases.

11 gallons per minute

(b) 65 gallons

Step-by-step explanation:

From inspection of the table, we can see that as time increases, the amount of water in the pool increases.

We are told that Ann adds water at a constant rate. Therefore, this can be modeled as a linear function.

The rate at which the water is increasing is the rate of change (which is also the slope of a linear function).

Choose 2 ordered pairs from the table:

$\textsf{let}\:(x_1,y_1)=(8, 153)$

$\textsf{let}\:(x_2,y_2)=(12,197)$

Input these into the slope formula:

$\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11$

Therefore, the rate at which the water in the pool is increasing is:

11 gallons per minute

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

$y-y_1=m(x-x_1)$