Help I need the intercept please.

A dvd is marked down 15% the regular price is $12 what is the sale price?

mash [69]

15% off means the dvd is selling for 85% of the original price ( 100% - 15%=85%)

Multiply the original price by 85%

12 x 0.85 = 10.20

The price is $10.20

4 0

2 years ago

Read 2 more answers

I need help finding what 875 divided by 46 is ?

Luda [366]

Answer:

19.02173913043478

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Word form for 18,429,000,050,000

Westkost [7]

Eighteen trillion four hundred twenty nine billion fifty thousand

5 0

2 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

In the diagram shown of circle A, tangent MB is drawn along with chords BAC and BF . Secant MFE intersects BAC at G. It is known

mote1985 [20]

Answer:

arc BF = 76°
∠M = 31°
∠BGE = 121°
∠MFB = 111°

Step-by-step explanation:

(a) ∠FBM is the complement of ∠FBC, so is ...

∠FBM = 90° -52° = 38°

The measure of arc BF is twice this angle, so is ...

arc BF = 2∠FBM = 2(38°)

arc BF = 76°

__

(b) ∠M is half the difference between the measures of arcs BE and BF, so is ...

∠M = (1/2)(138° -76°) = 62°/2

∠M = 31°

__

(c) arc FC is the supplement to arc BF, so has measure ...

arc FC = 180° -arc BF = 180° -76° = 104°

∠BGE is half the sum of arcs BE and FC, so is ...

∠BGE = (1/2)(arc BE +arc FC) = (138° +104°)/2

∠BGE = 121°

__

(d) ∠MFB is the remaining angle in ∆MFB, so has measure ...

∠MFB = 180° -∠M -∠FBM = 180° -31° -38°

∠MFB = 111°

3 0

3 years ago