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Romashka [77]
2 years ago
11

Please help me with this

Mathematics
1 answer:
PSYCHO15rus [73]2 years ago
6 0

Answer:

Step-by-step explanation:

Both angles added together = 180 degrees. They both have a straight line as their base.

So let <ABC = x

Then <CBD = 3x      Add the two angles

x + 3x = 180              Combine

4x = 180                    Divide by 4

4x/4 = 180/4

x = 45

So 3x = 3*45 = 135

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Y=6x 2x+3y=-20. (Solving by substitution)
almond37 [142]
Y=6x, 2x+3y=-20
Substitution means just plugging in another equation equal to a single variable in order to solve for the other variable, and then the first. In this case, the equation being plugged in is y=6x, so 6x will be put in place of y in the equation 2x+3y=-20.
2x+3(6x)=-20
2x+18x=-20
20x=-20
x=-1
And then x=-1 can be plugged into the original equation to find y. (And, if you wish, into the y= equation to check it.)
2(-1)+3y=-20
-2+3y=-20
3y=-20+2
3y=-18
y=-6
And to check it, we plug the two into the original y= equation.
y=6x
-6=6(-1)
-6=-6
They are equal, so both check out.
x=-1, y=-6
7 0
3 years ago
Read 2 more answers
Solve the equation 14=-(x-8)
sashaice [31]

Answer:

x = -6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
For for x.Help plz:) -2x+6=10
Sphinxa [80]

Answer:

x=-2

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

−2x+6−6=10−6

−2x=4

Step 2: Divide both sides by -2.

−2x=4

7 0
3 years ago
Find the minimum and maximum of f(x,y,z)=x2+y2+z2 subject to two constraints, x+2y+z=7 and x−y=6.
sweet-ann [11.9K]
Use the method of Lagrange multipliers. We have Lagrangian

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-7)+\lambda_2(x-y-6)

with partial derivatives (set equal to 0) of

L_x=2x+\lambda_1+\lambda_2=0
L_y=2y+2\lambda_1-\lambda_2=0
L_z=2z+\lambda_1=0
L_{\lambda_1}=x+2y+z-7=0
L_{\lambda_2}=x-y-6=0

As x+2y+z=7, and x-y=6, we can obtain

\dfrac12L_x+L_y+\dfrac12L_z=0\implies3\lambda_1-\dfrac12\lambda_2=-7
L_x-L_y=0\implies\lambda_1-2\lambda_2=12
\begin{cases}3\lambda_1-\frac12\lambda_2=-7\\\lambda_1-2\lambda_2=12\end{cases}\implies\lambda_1=-\dfrac{40}{11},\lambda_2=-\dfrac{86}{11}

From this, we find a single critical point:

2x-\dfrac{40}{11}-\dfrac{86}{11}=0\implies x=\dfrac{63}{11}
\dfrac{63}{11}-y=6\implies y=-\dfrac3{11}
\dfrac{63}{11}-\dfrac6{11}+z=7\implies z=\dfrac{20}{11}

At this point, we have a value of

f\left(\dfrac{63}{11},-\dfrac3{11},\dfrac{20}{11}\right)=\dfrac{398}{11}

To determine what kind of extremum occurs at this point, we check the Hessian of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

We observe that \det\mathbf H(x,y,z)=8>0 at any point (x,y,z), and that the eigenvalues of this matrix are all positive (2 with multiplicity 3), so \mathbf H is positive definite. By the second partial derivative test, this means f(x,y,z) attains a minimum at this critical point. Meanwhile, f has no maximum value.
5 0
3 years ago
Can someone solve this for me?
AVprozaik [17]

Answer:

The slope is 7

Step-by-step explanation:

The number by X

5 0
4 years ago
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