Ask question

Ask question

All categories

2 years ago

12

Pls, I request ya all, exams are near, I am not able to figure out the answer (along with process) of this question. Pls, I will

mark as brainliest tose who answer me with the correct process.
Thank You

2 answers:

sukhopar [10]2 years ago

8 0

Answer:

a was i believe 10 degrees and b dropped 3- degrees

Step-by-step explanation:

Because 4- 3- 2- 1- 0 1 2 3 4 5

1 2 3 4 5 6 7 8 9

4 is the start don't forget that

And for b 5 dropped from 7 degrees just take 5 now ur at 0 and u have 3 more to take boom ur at 3- hope this hellped :)

DENIUS [597]2 years ago

6 0

What this question is asking us is

A) From -40C to 5 0c How much did it rise

The formula for this would be

Risen Temperature - Initial Temperature

The risen temparauture is 5 so we plug the 5 in first then the 4 so it would look like

5-(-4)=9

And remeber when you have 2 negatives it becomes a positie, 5+4=9

The asnwer for A is 9 degress

B) The tempurater then fell by 7 degress from 5 so we just have to do

5 - 7

which is -2 Celcius

You might be interested in

Dillon can buy a fixed-rate bond with a 4% coupon, a $10,000 principal, and five years left to maturity or a TIPS with the same

Keith_Richards [23]

<span>3 Both yields are roughly equivalent.</span>

7 0

3 years ago

Read 2 more answers

Given parallelogram ABCD , diagonals AC and BD intersect at point E. AE=2x, BE=y+10, CE=x+2 and DE=4y−8 . Find the length of AC

Crank

Answer:

I think the answer shod be B

3 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

What is 7.39 rounded to the nearest whole number?

lutik1710 [3]

7 is the answer 4 and lower you round down 5 and up is up

5 0

3 years ago

Read 2 more answers

The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by: P(x)=-12x^2 +12

Minchanka [31]

Given:

The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by:

$P(x)=-12x^2+120x$

To find:

The current that will produce the maximum power.

Solution:

We have,

$P(x)=-12x^2+120x$

Here, leading coefficient is negative. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

If a parabola is $f(x)=ax^2+bx+c$ , then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given function, a=-12 and b=120. So,

$-\dfrac{b}{2a}=-\dfrac{120}{2(-12)}$

$-\dfrac{b}{2a}=-\dfrac{120}{-24}$

$-\dfrac{b}{2a}=5$

Putting x=5 in the given function, we get

$P(5)=-12(5)^2+120(5)$

$P(5)=-12(25)+600$

$P(5)=-300+600$

$P(5)=300$

Therefore, 5 watt current will produce the maximum power of 300 amperes.

6 0

3 years ago