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3 years ago

5

Help with question three (the photo above) Pleaseee helpp

1 answer:

Pie3 years ago

8 0

The answer would
Be 32.5%

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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

Read 2 more answers

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

3 years ago

Look at the graph.

Rainbow [258]

Answer:

$f(x) =80 \cdot \: {( \frac{1}{4} )}^{x - 1}$

Step-by-step explanation:

Fro the graph, the first three terms of the sequence are 80,20,5,...

This is a geometric sequence with first term

$f(1)= 80$

The common ratio is

$r = \frac{20}{80} = \frac{5}{20} = \frac{1}{4}$

The function is given by

$f(x) = f(1) \cdot \: {r}^{x - 1}$

We substitute the values we got from the graph into this formula to get:

$f(x) =80 \cdot \: {( \frac{1}{4} )}^{x - 1}$

The correct answer is A

5 0

3 years ago

Read 2 more answers

What is 6/13 in short decimal form.

elena55 [62]

I think I'll be 6/13= 0.4615

6 0

4 years ago

Read 2 more answers

HELP PLEASE DUE IN 3 MINUTES

romanna [79]

Answer:

It's 143

Step-by-step explanation:

Your welcome have a good day!!!

4 0

3 years ago