How to solve it? And how to describe it better ?

Find sin A. 12/13 B. 1 C. 13/12 D. 13/5

inn [45]

Answer:A 12/13

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Someone help solve 6 & 7 (will give brainliest)

riadik2000 [5.3K]

6.
$T(I(h))= \sqrt{5000( \frac{12h-h^2}{36} )}$

Since, function I stated that hours, h, pertains to the hours after 6am, and the question asks what is the temperature by 2pm, we will have h=8 because 2pm is 8 hours after 6am.

So, $T(I(h))= \sqrt{5000( \frac{12(8)-(8)^2}{36} )}

T(I(h))= \sqrt{5000( \frac{96-64}{36} )}

T(I(h))= \sqrt{5000( \frac{32}{36} )}

T(I(h))= \sqrt{ \frac{40000}{9} }

T(I(h))=66.6666 or 67$

so the answer is (3), 67.

6 0

3 years ago

You roll a twelve-sided die (having values one through twelve on its faces). What is the probability that the value of the roll

vlada-n [284]

Step-by-step explanation:

Probability

Probability is a measure of the likelihood that an event will happen.

When dealing with probability, the outcomes of a process are the possible results. For example, when a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, and 6. In mathematical language, an event is a set of outcomes, which describe what outcomes correspond to the "event" happening. For instance, "rolling an even number" is an event that corresponds to the set of outcomes {2, 4, 6}. The probability of an event, like rolling an even number, is the number of outcomes that constitute the event divided by the total number of possible outcomes. We call the outcomes in an event its "favorable outcomes".

Probability =

6 0

3 years ago

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

allsm [11]

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

or

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

7 0

3 years ago

F is a twice differentiable function that is defined for all reals. The value of f "(x) is given for several values of x in the

nadezda [96]

The correct answer is actually the last one.

The second derivative $f''(x)$ gives us information about the concavity of a function: if $f''(x)$ then the function is concave downwards in that point, whereas if $f''(x)>0$ then the function is concave upwards in that point.

This already shows why the first option is wrong - if the function was concave downwards for all x, then the second derivate would have been negative for all x, which isn't the case, because we have, for example, $f''(8)=5$

Also, the second derivative gives no information about specific points of the function. Suppose, in fact, that $f(x)$ passes through the origin, so $f(0)=0$ . Now translate the function upwards, for example. we have that $f(x)+k$ doesn't pass through the origin, but the second derivative is always $f''(x)$ . So, the second option is wrong as well.

Now, about the last two. The answer you chose would be correct if the exercise was about the first derivative $f'(x)$ . In fact, the first derivative gives information about the increasing or decreasing behaviour of the function - positive and negative derivative, respectively. So, if the first derivative is negative before a certain point and positive after that point. It means that the function is decreasing before that point, and increasing after. So, that point is a relative minimum.

But in this exercise we're dealing with second derivative, so we don't have information about the increasing/decreasing behaviour. Instead, we know that the second derivative is negative before zero - which means that the function is concave downwards before zero - and positive after zero - which means that the function is concave upwards after zero.

A point where the function changes its concavity is called a point of inflection, which is the correct answer.

7 0

3 years ago