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aev [14]
3 years ago
5

I NEED A PROFESSIONAL ON MATH PROBLEMS NOW!!

Mathematics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

$62

Step-by-step explanation:

hope this helped!

p.s it would be cool if you gave me brainliest.

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Please help me solve for13 and 14
deff fn [24]

Answer:

13. A

14. True

Step-by-step explanation:

The months cancel out because they are the same unit in the right positions, and all you have to do for the second one is multiply

4 0
3 years ago
Solve for x: 5x+ 100 3x + 112
Anna71 [15]

Answer:

you can't solve if it does not equal anything, but you can simplify

Step-by-step explanation:

simplified it is,

8x+212

7 0
3 years ago
Y=2x+1<br>y=4x-5<br><br>please help​
levacccp [35]

The point of intersection would be (3, 7)

8 0
3 years ago
Find the total surface area of the following pyramids: A regular square pyramid has a height of 4 and a base with area 36.
Artemon [7]

Answer:

2623.6144m²

Step-by-step explanation:

Total Surface Area of a Square Pyramid

a * (a + √(a² + 4h²))

where,

a = area of base

h = height

Using the formula above

a * (a + √(a² + 4h²))

data given :

a = 36

h = 4

a * (a + √(a² + 4h²))

T.S.A. = a * (a + √(a² + 4h²))

Inputting values

T.S.A. = 36 * (36 + √(36² + 4 * 4²))

T.S.A. = 36 * (36 + √(36² + 4 * 16))

T.S.A. = 36 * (36 + √(36² + 64))

T.S.A. = 36 * (36 + √(1296 + 64))

T.S.A. = 36 * (36 + √(1360))

T.S.A. = 36 * (36 + 36.87817782917154)

T.S.A. = 36 * (72.87817782917154)

T.S.A. = 36 * 72.8782

T.S.A. = 2623.614401850175

Therefore Total Surface Area (T.S.A.) is 2623.6144m²

6 0
3 years ago
Read 2 more answers
Solve the initial value problem y′+y=f(t),y(0)=0 where f(t)={1,−1, if t&lt;4 if t≥4 Use h(t−a) for the Heaviside function shifte
Anton [14]

Looks like the function on the right hand side is

f(t)=\begin{cases}1&\text{for }t

We can write it in terms of the Heaviside function,

h(t-a)=\begin{cases}1&\text{for }t\ge a\\0&\text{for }t>a\end{cases}

as

f(t)=h(t)-2h(t-4)

Now for the ODE: take the Laplace transform of both sides:

y'(t)+y(t)=f(t)

\implies s Y(s)-y(0)+Y(s)=\dfrac{1-2e^{-4s}}s

Solve for <em>Y</em>(<em>s</em>), then take the inverse transform to solve for <em>y</em>(<em>t</em>):

(s+1)Y(s)=\dfrac{1-e^{-4s}}s

Y(s)=\dfrac{1-e^{-4s}}{s(s+1)}

Y(s)=(1-e^{-4s})\left(\dfrac1s-\dfrac1{s+1}\right)

Y(s)=\dfrac1s-\dfrac{e^{-4s}}s-\dfrac1{s+1}+\dfrac{e^{-4s}}{s+1}

\implies y(t)=1-h(t-4)-e^{-t}+e^{-(t-4)}h(t-4)

\boxed{y(t)=1-e^{-t}-h(t-4)(1-e^{-(t-4)})}

4 0
3 years ago
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