Line a in parallel to b, m

6/9 is a rational number. Select all the rational numbers that are equivalent to 6/9

timama [110]

2/3 and 0.667 are rational numbers equivalent to 6/9

3 0

3 years ago

Read 2 more answers

What’s a matrix and determinant?

Margaret [11]

Answer:

yan na po

Step-by-step explanation:

hope it helps po

7 0

3 years ago

PLZ help now i need help n

vichka [17]

Answer:

7

Step-by-step explanation:

hey there,

< Here's how you do subtraction in fractions!

Make sure you find a least common denominator. Here, it's given to you to be 15. Usually, you can just multiply the two denominators but then it would be quite large at times, so you kind of just have to guess.

How did they go from $\frac{4}{5} x-\frac{1}{3} x$ to $\frac{?}{15} x$ ?

Well, they used a least common denominator. 5 was multiplied by 3 to create 15, and 3 was multiplied by 5. But when you multiply the bottoms, you have to multiply the tops of the fraction too (the numerators).

So 4/5x was multiplied by 3. Denominator turned out to be 15, and now the top is 4x3 = 12. So the fraction is 12/15.

1/3x, on the other hand, was multiplied by 5. Denominator was also 15 so they're the same, and top is 1x5 = 5. So the fraction is 5/15.

Now, let's put them together.

$\frac{12}{15}x-\frac{5}{15} x$

When subtracting x, the x just stays there, so if it was 10x-4x, the answer would've been 6x. So now let's subtract!

12-5 = 7, and the denominator stays the same.

The answer is $\frac{7}{15} x$ . So the numerator, or answer to your question, would be 7.

Hope this helped! Feel free to ask anything else.

5 0

3 years ago

Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0

Snowcat [4.5K]

Answer:

y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = $e^{-cs}$

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)} − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 5s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 3s + 2s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[(s + 2)(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒Y(s) = $\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}$

Now,

Let

$\frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)$

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

$\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}$

So,

Y(s) = $3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}]$

⇒Y(s) = $3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}$

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ $e^{-2(t-2)} - e^{-5(t - 2)}$ ] - 4u₅(t) [ $e^{-2(t-5)} - e^{-5(t - 5)}$ ]

⇒y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

It is the required solution.

3 0

3 years ago

A right triangle has a hypotenuse of 10 mi. and a leg of 6 mi., what is the length of the other leg?

slava [35]

Answer:

8mi

Step-by-step explanation:

Using Pythagoras theorem

Hypotenuse square = opposite square + adjacent square.

Let the other leg be x

10 square = 6 square + x square

100 = 36 + x square

Collect like terms

X square = 100 -36

X square = 64

Find square root of both sides

X = square root of 64

X =8mi

I hope this was helpful, Please mark as brainliest

7 0

3 years ago