Find the value of x.<br> 5x° 130°<br><br> ASAP

Consider the system of equations.

Liula [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

The surface area of a right cone is 400.2 m2. The radius of the cone is 6.0 m. Determine the height of the cone to the nearest m

svetoff [14.1K]

S = πr(r + √(h² + r²))
  400.2 = 3.14(6)(6 + √(h² + 6²))
  400.2 = 18.84(6 + √(h² + 36))
  18.84   18.84
  21¹⁰⁹/₄₇₁ = 6 + √(h² + 36))
  - 6 - 6
  15¹⁰⁹/₄₇₁ = √(h² + 36)
231²²¹⁰⁰⁵/₂₂₁₈₄₁ = h² + 36
- 36   - 36
195²²¹⁰⁰⁵/₂₂₁₈₄₁ = h²
  14 ≈ h

8 0

3 years ago

In a certain state, license plates each consist of 2 letters followed by either 3 or 4 digits. How many differen license plates

coldgirl [10]

Answer:

26 × 26 × 10 × 10 × 10 = 676 , 000 possibilities

Step-by-step explanation:

There is nothing stating that the letters and numbers can't be repeated, so all 26 letters of the alphabet and all 10

digits can be used again.

If the first is A, we have 26 possibilities:

AA, AB, AC,AD,AE ...................................... AW, AX, AY, AZ.

If the first is B, we have 26 possibilities:

BA, BB, BC, BD, BE .........................................BW, BX,BY,BZ

And so on for every letter of the alphabet. There are 26 choices for the first letter and 26 choices for the second letter. The number of different combinations of 2 letters is: 26 × 26 = 676

The same applies for the three digits. There are 10 choices for the first, 10

for the second and 10 for the third:

10 × 10 × 10 = 1000

So for a license plate which has 2 letters and 3 digits, there are: 26 × 26 × 10 × 10 × 10 = 676 , 000 possibilities.

Hope this helps.

8 0

3 years ago

A coin is tossed 50 times. It lands on heads 30 times. Based on this outcome, how many times is the coin expected to land on hea

BaLLatris [955]

Answer:

D: 120

Step-by-step explanation:

30/50 = 120/200

3 0

3 years ago

Read 2 more answers

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

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