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quester [9]
3 years ago
12

(7x^6+2x^5-3x^2+9x)+(5x^5+8^3-6x^2+2x-5)

Mathematics
1 answer:
Reil [10]3 years ago
7 0

Answer:

Step-by-step explanation:

Combine like terms

7x^6

2x^5+5x^5= 7x^5

8x^3

-3x^2-6x^2= -9x^2

9x+2x= 11x

-5

Put back into order

7x^6+7x^5+8x^3-9x^2+11x-5

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Please this is urgent I need help please answer me
Verizon [17]

Answer:

u have the answer yet?

Step-by-step explanation:

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8 0
3 years ago
Solve the system of equations
rewona [7]

Answer:

x = 1/4

y = -1/2

z = 9/4

Step-by-step explanation:

Here we have a system of 3 equations with 3 variables:

4*x + 2*y + 1 = 1

2*x - y = 1

x + 3*y + z = 1

The first step to solve this, is to isolate one of the variables in one of the equations, let's isolate "y" in the second equation:

2*x - y = 1

2*x - 1 = y

Now that we have an expression equivalent to "y", we can replace this in the other two equations:

4*x + 2*(2*x - 1) + 1 = 1

x + 3*(2*x - 1) + z = 1

Now let's simplify these two equations:

8*x - 1 = 1

7*x - 3 + z = 1

Now, in the first equation we have only the variable x, so we can solve that equation to find the value of x:

8*x - 1 = 1

8*x = 1 + 1 = 2

x = 2/8 = 1/4

Now that we know the value of x, we can replace this in the other equation to find the value of z.

7*(1/4) -3 + z = 1

7/4 - 3 + z = 1

z = 1 + 3 - 7/4

z = 4 - 7/4

z = 16/4 - 7/4 = 9/4

z = 9/4

Now we can use the equation y = 2*x - 1 and the value of x to find the value of y:

y = 2*(1/4) - 1

y = 2/4 - 1

y = 1/2 - 1

y = -1/2

Then the solution is:

x = 1/4

y = -1/2

z = 9/4

7 0
3 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2
Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either   \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t  = e^−2(0)  = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt}  = -2te^{-2t}  = -2(0)e^{-2(0)}  = 0

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0
3 years ago
At noon, a submarine was located at 5 12 meters below
Cerrena [4.2K]
-512 is the answer because 512 is a negative solution in Integers math
7 0
3 years ago
Which ordered pair is a solution to this equation? 3x + 7y = 13 (4, 1) (3, 1) (13, 2) (2, 1)
MAVERICK [17]
<span>x = 2, y = 1
       
= 3*2 + 7*1
   
= 6+7
   
= 13
       
Hence, (2, 1) is the answer.</span>
8 0
3 years ago
Read 2 more answers
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