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3 years ago

Define Hypotenus I have forgotten what it is

2 answers:

4 0

In geometry, a hypotenuse is the longest side of a right-angled triangle, the side opposite the right angle. The length of the hypotenuse can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides.

<em>It is useful in different situations involving the need to find distance or the measure of an angle.</em>

Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Take a square root of sum of squares: c = √(a² + b²)

irinina [24]3 years ago

3 0

A^2 + b^2 = c^2. A and B are the legs and C is the hypotenuse. fill in the leg lengths where they belong then solve for C.

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Round 2.408 to the nearest hundred

kaheart [24]

2.408 Rounded to the nearest hundredth's place is 2.41

4 0

3 years ago

Read 2 more answers

2/9 g =-8

Rasek [7]

Answer:

-8

Step-by-step explanation:

7 0

2 years ago

Which is the solution to the inequality?

makvit [3.9K]

Answer:

b>3.133333333

Step-by-step explanation:

13/5<b-8/15

13/5+8/15<b

39+8/15<b

47/15<b

3.13333333<b

b>3.13333333

6 0

3 years ago

The cost, in millions of dollars, for a company to manufacture x thousand speed boats is given by the function C(x) = 3x2 - 30x

nalin [4]

Answer:

5000

Step-by-step explanation:

In vertex form, the cost function is ...

C(x) = 3(x² -10x) +175 = 3(x² -10x +25) +175 -3(25)

C(x) = 3(x -5)² +100

The vertex is at x=5, so the cost function will be minimized when 5000 speed boats are produced.

7 0

4 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago