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Tresset [83]
3 years ago
14

Answer to the question

Mathematics
1 answer:
natta225 [31]3 years ago
6 0

Answer:

Value of x = -1 and y = -4

Step-by-step explanation:

BC = 3 (given)

BC = AD ( opposite sides of a parallelogram are equal)

∴ AD = 3

AD = x - y (given)

∴ 3 = x - y

3 + y = x -------- (1)

CD = 2 (given)

CD = AB (opposite sides of a parallelogram are equal)

∴ AB = 2

AB = 2x - y (given)

∴ 2 = 2x - y ----------- (2)

Putting the value of x in (1) To (2)

2 = 2(3+y) - y

2 = 6 + 2y - y

2 = 6 + y

2 - 6 = y

-4 = y

Now, substitute the value of y in (1)

x = 3 + y

x = 3 + (-4)

x = 3 - 4

x = -1

Therefore, the value of x = -1 and y = -4.

<u>Cross-checking</u>

2x - y = 2

2*-1 - (-4) = 2

-2 -(-4) = 2

-2 + 4 = 2

2 = 2

LHS = RHS

x - y = 3

-1 -(-4) = 3

-1+4 = 3

3 = 3

LHS = RHS

Thus verified

Hope u understand

Please mark as brainliest

Thank You

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What is the solution of the inequality below? <br> y-1≥0
irinina [24]

Answer:

y≥1

Step-by-step explanation:

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y-1≥0

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3 years ago
Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.
gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

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Step-by-step explanation:

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Please have a look at the attached photo.

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