What is the domain and range of the following:

Brooke has to set up 70chairs in equal rows for the class talent show.But,there is not room for more than 20 rows. What are the

Zinaida [17]

Answer:

R = rows

C = columns

The chairs will be arranged in a way such that R * C $\geq$ 70

We know that R $\leq$ 20

70 / 20 = 3.5; however, you can't have half of a chair.

So find all the factors of 70 which don't exceed 20.

1, 70

2, 35

<h2>5, 14</h2><h2>7, 10</h2><h2></h2><h2>There are four ways that Brooke can set up the chairs:</h2><h2>5 rows of 14</h2><h2>7 rows of 10</h2><h2>10 rows of 7</h2><h2>14 rows of 5</h2>

3 0

2 years ago

What is the value of x?<br><br><br><br> Enter your answer in the box.<br><br> x =

polet [3.4K]

The two angles are the same value. To find, think of the equation 3<em>x</em>+50=6<em>x</em>-10. Put the 6<em>x</em> to the other side of the equation. You shall get -3x+50=-10. Move the 50 to the other side, and you'll end up with -3<em>x</em>=-60. Divide each number by -3 to find <em>x</em> now. Your final answer will be <em>x=20</em>.

Answer: <em>x=20</em>

8 0

2 years ago

Solve the differential equation.<br><br> (y^2 + xy^2)y' = 1

stira [4]

$(y^2 + xy^2)y' = 1 \ \Rightarrow\ (y^2 + xy^2)\frac{dy}{dx}= 1 \ \Rightarrow\\ \\
y^2(1 + x) \frac{dy}{dx} = 1 \ \Rightarrow\ y^2 dx = \frac{dx}{1 + x}\ \Rightarrow \\ \\
\int y^2 dx =\int \frac{dx}{1 + x} \ \Rightarrow\ \frac{1}{3}y^3 = \ln|1+x| + C\ \Rightarrow \\ \\
y^3 = 3 \ln|1 + x| + 3C\ \Rightarrow\ y = \sqrt[3]{3\ln|1+x| + K},\ \text{where $K = 3C$}
$

6 0

2 years ago

dillon pays x dollars to get a haircut. he gives the stylist a 15% tip. the expression representing his total cost is x + 0.15x

bagirrra123 [75]

Answer:

B) 1.15x, adding 15% to the cost of haircut is same as to multiply cost by 1.15

Step-by-step explanation:

Given expression = x + 15% of x

x + (15/100) x = x + 0.15x

= (1+ 0.15) x = 1.15x

15% = 0.15 , added to x = 1.15x

4 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

2 years ago