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2 years ago

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Tatiana practiced a total of 12 hours last week to prepare for her gymnastics meet. The line plot shows how long she practiced o

n the first six days of the week. If Tatiana practiced every day last week, how long did she practice on the last day?

1 answer:

Free_Kalibri [48]2 years ago

3 0

Answer:

nigrto

Step-by-step explanation:

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2 years ago

How would I solve this equation for mass (m)?

zaharov [31]

Simple...

you have: $F= \frac{G m^{2} }{ d^{2} }$

You want to find m-->>>

1.)Isolate m

$d^{2} *F= \frac{G m^{2} }{ d^{2} } * d^{2}$

Leaving you with...

F* $d^{2}$ = $G m^{2}$

Keep isolating m-->>>

$F* d^{2} =G m^{2}$

$\frac{F* d^{2} }{G} = \frac{G m^{2} }{G}$

Square root to solve what just m is-->>>

$\sqrt{ \frac{F* d^{2} }{G} } = \sqrt{ m^{2} }$

= $\sqrt{ \frac{F* d^{2} }{G} } =m$

That is how you solve for m...

Thus, your answer.

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2 years ago

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2 years ago

Ben wrestles in the 132 to 138- pound weight class. If he currently weighs 133 pounds, how much can he gain if he wishes to stay

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Step-by-step explanation:

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2 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

a)

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

b)

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

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

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

3 years ago