Make y the subject of the formula:A = xy + yz

larisa86 [58]

Answer: $0 \ \textless \ x \le 8$

What is given to you is interval notation. We have the interval start at 0 and end at 8. The value 0 is not included in the interval as indicated by the parenthesis. So we go with a "less than" sign (instead of a "less than or equal to" sign)

The value 8 is included since a square bracket is used here. The use of "or equal to" is needed to make sure the endpoint 8 is included.

So x can be any number between 0 and 8. It can't be 0 but it can be 8.

3 0

3 years ago

The perimeter of a triangle is 32 feet. One side of the triangle is 3 times the second side. The third side is 12 feet longer

Ivenika [448]

We have a triangle whose perimeter is 32 feet. Some information regarding the sides of the triangle is given:

GiveN:

One side is 3 times the second side.
Third side is 12 feet longer than the second side.

As we can see that, two sides of the triangle are defined the second sides. It means First and third side can be expressed in the form of second side.

So let the second side be x

Then,

First side = 3x
And, third side = x + 12

We know that, perimeters is the sum of the lengths of all sides of the triangle, So it can be written as:

$3x + x + x + 12 = 32$

Solving it further,

$5x + 12 = 32$

Subtracting 12 from both sides,

$5x + 12 - 12 = 32 - 12$

$5x = 20$

Dividing 5 from both sides,

$\frac{5x}{5} = \frac{20}{5}$

$x = 4$

Then,

First side = 3x = 12 feet
Second side = 4 feet
Third side = x + 12 = 16 feet

And we are done with the answer !!

#CarryOnLearning

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4 0

3 years ago

What is the area of this quadrilateral?

Sever21 [200]

24 square feet is the answer

7 0

3 years ago

Did you hear about the two punsters who told a lot of jokes about cats

Serggg [28]

Ooo. I I didn't hear please do tell

4 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