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Sedbober [7]
3 years ago
11

Find the length of the missing side, x=

Mathematics
1 answer:
Paladinen [302]3 years ago
8 0

{x}^{2}  +  {11.9}^{2}  =   {14.7}^{2}  \\ x =  \sqrt{ {14.7}^{2}  -  {11.9}^{2} }  \\  x =  \sqrt{74.48}  \\  x =  8.6301796042...

Answer: 9 or 8.6

  • I don't know, need I round to 8.6 or 9, so you can try both answers
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19. Find the measure of angle ABC.
aliya0001 [1]

Answer:

72

Step-by-step explanation:

6x+6x+8x=180 (∠ sum of Δ)

20x=180

x=9

∠ABC=8(9)

         =72

8 0
2 years ago
I want to know the answer of this question
Molodets [167]
\bf \textit{circumference of a circle}\\\\
C=2\pi r\qquad 
\begin{cases}
r=radius\\
-----\\
r=5x^2
\end{cases}\implies C=2\pi 5x^2\implies C=10\pi  x^2
\\\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad 
\begin{cases}
r=radius\\
-----\\
r=5x^2
\end{cases}\implies A=\pi (5x^2)^2\implies A=\pi (5^2x^{2\cdot 2})
\\\\\\
A=\pi (25x^4)\implies A=25\pi x^4
4 0
3 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
3 years ago
Help me plzzzzzzzzzzzz
Paraphin [41]

Answer: -27/7

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Simplify sin(3pi minus x)
Slav-nsk [51]

Answer:

sinx

Step-by-step explanation:

hello :

note :

for all x in ℝ :   1) sin(x+2π) = sinx        2)  sin(π-x) = sinx

in this exercice :

3π - x = 2π+(π - x)

sin(3π - x) = sin(2π +(π - x)) =sin(π-x) = sinx

8 0
2 years ago
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