How do I solve this. I have a test tomorrow and I don't know how to solve this

Mathematics

1 answer:

joja [24]3 years ago

8 0

I already answered this for you its 195

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Does anyone know this?

barxatty [35]

Answer:

$\Huge \boxed{-2}$

Step-by-step explanation:

$f(x)=-2x^2-2x+10$

For f(2), the input of the function is 2.

Let x = 2.

$f(2)=-2(2)^2-2(2)+10$

Evaluate.

$f(2)=-2(4)-2(2)+10$

$f(2)=-8-4+10$

$f(2)=-2$

8 0

4 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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4 years ago

Alana and Ramiro are writing an article for the school newspaper about whether their classmates prefer to buy their lunch or bri

MrRissso [65]

Answer:Ramiro’s method is

Step-by-step explanation:

Did the assignment

3 0

3 years ago

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Simplify the expression:<br> sinx+cosx/sinxcosx

Keith_Richards [23]