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Flauer [41]
3 years ago
7

PLEASE ANSWERRRRRRRRR Solve the system of equations.

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
6 0

Answer:

x=6 y=-3    (6,-3)

Step-by-step explanation

You might be interested in
Match each radius or diameter of a circle with its approximate circumference.
jok3333 [9.3K]

Answer:

radius: 3 units <----> Circumference: 18.84 units

Diameter: 12 units <----> Circumference: 37.68 units

Diameter: 36 units <----> Circumference: 113.04 units

Radios: 5 unites <----> Circumference: 31.4 units

Step-by-step explanation:

How I got the answers for the radiuses:

3 (this is only half the diameter so if you do 3 x 2 you will get the diameter.)

3 x 2 = 6

C = pi x d

C = 3.14 x 6

C = 18.84 [this is the first radios]

5 (this is only half the diameter so if you do 5 x 2 you will get the diameter.)

5 x 2 = 10

C = pi x d

C = 3.14 x 10

C = 31.4 [this is the second radios]

How I got the diameters:

12 (this is the full diameter so we do not have to multiply it by 2)

C = pi x d

C = 3.14 x 12

C = 37.68 [this is the first Diameter]

36 (this is the full diameter so we do not have to multiply it by 2)

C = pi x d

C = 3.14 x 36

C = 113.04 [this is the second Diameter]

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5 0
3 years ago
42 divided by 4 and 67 divided by8
Bezzdna [24]
10.5 and 8.4 ......................,,,,,
6 0
3 years ago
Read 2 more answers
Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/
devlian [24]

u(x,y,z)=x^2+yz

\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}

At the point (p,r,\theta)=(2,2,0), we have

\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}

Denote by f_x:=\dfrac{\partial f}{\partial x} the partial derivative of a function f with respect to the variable x. We have

\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}

The Jacobian is

\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}

By the chain rule,

u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y

u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}

u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y

u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}

u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta

u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}

7 0
3 years ago
Solve - 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] + 3 ( x - 1 ) (and could you please explain it to me?? I really don't get it.
Mazyrski [523]

Answer:

Step-by-step explanation:

(x+2)(x-1)= x^2+2x-x-2= x^2 +x -2\\\\( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 =x^2+x-2-x^2+1=x-1\\\\- 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] =-3(x-1)=-3x+3\\\\- 3 [ ( x + 2 ) ( x - 1 ) - x ^ { 2 } + 1 ] + 3 ( x - 1 ) =-3x+3+3x-3=6x\\

5 0
3 years ago
what is 5 divide by zero? but the answer is not zero???????????????????????????????????????????????????????????????????????????
krek1111 [17]
That's right.  The answer is not zero.  In fact, there's no answer at all,
simply because there's no question.  In math in general, division by zero
is "not permitted", and if people start talking about it, the answer is called
"indeterminate" ... that means vague, and can't be determined.

Division is repeated subtraction.  6 divided by 2 means:  "How many times
can you take 2 away from 6 before the 6 is all used up and gone ?", and the
answer is: 3 times and then it's gone. 

So the question is really asking:  "How many times can you take zero away
from 5 before the 5 is all used up and gone ?".  There's no answer, because
no matter how many times you take away zero, you can never use up the 5 .

One more comment on the subject:  Save your question marks.  One is plenty.

8 0
3 years ago
Read 2 more answers
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