the lenght of a rope is 2.589 yards round the lenght to the nearest tenth of a yard.2.589 yards rounds to________

<img src="https://tex.z-dn.net/?f=%5Chuge%5Ccolor%7Byellow%7D%5Cunderline%20%5Ccolorbox%7Bviolet%7D%20%7B%5Ccolor%7Bred%7D%7BQUE

Paladinen [302]

Answer:

FALSE
TRUE
TRUE

Step-by-step explanation:

1.

Non-standard units can be used to measure volume
Examples : m³, in³, cm³, mm³, L

2.

The volume of a rectangular prism can be found by multiplying the length, width, and height
V = lwh

3.

If there are smaller units, then naturally more of the units will be used
If there are larger units, then less of the units will be used

8 0

2 years ago

Please heeelp !! Missing triangles !!

Airida [17]

Answer: ABC seem to be equilateral

so A=B=C=60

D=42

E=35

F=35

G=55

H=45

I=45

j=15

k=15

L=75

M=180

N=11

O=35

P=11

Q=55

R=55

S=32

T=32

U=69

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please Help! The graph of y=f(x) is shown below. What are all of the real solutions of f(x)=0? (Picture of graph included)

qwelly [4]

Answer:

(7, 0), (-8, 0), (0, 0)

Step-by-step explanation:

y = f(x) = 0

the real solutions are the ones that cross the x-axis because when a point is on the x-axis, its y coordinate will be 0.

=> (7, 0), (-8, 0), (0, 0)

6 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

Which of the following are opposite rays

maks197457 [2]

The answer is letter i

3 0

3 years ago

the lenght of a rope is 2.589 yards round the lenght to the nearest tenth of a yard.2.589 yards rounds to_________yards.