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3 years ago

Choose the expression that is equivalent to both −x/y and x/−y.

Mathematics

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

Answer:

1÷y/-x

Step-by-step explanation:

1÷y/-x

1×-x/y

-x/y

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PLZ HELP SCALE FACTOR TRIANGLE<br><br> A. 10.5 cm<br> B. 31.5 cm<br> C. 63 cm<br> D. 189 cm

Anarel [89]

If two similar triangles have sides in the ratio a : b, then their areas are in the ratio a² : b².

We have the ratio:

$54:18=\dfrac{54}{18}=3:1$

Area of the smaler triangle = x

Area of the larger triangle = 567 cm²

Therefore we have the equation:

$567:x=3^2:1^2\\\\\dfrac{567}{x}=\dfrac{9}{1}\qquad|\text{cross multiply}\\\\9x=567\qquad|\text{divide both sides by 9}\\\\x=63$

<h3>Answer: C. 63 cm²</h3>

3 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1

6 0

3 years ago

Can somebody help me with this math problem?

ankoles [38]

She needs 0.1 because 4x4 is 16. :)

7 0

3 years ago

An engineer is designing a storage compartment in a spacecraft .The compartment must be 2 meters longer than it is wide, and its

serg [7]

Answer:

Let x be the width of the compartment,

Then according to the question,

The length of the compartment = x + 2

And the depth of the compartment = x -1

Thus, the volume of the compartment, $V = (x+2)x(x-1) = x^3 + x^2 - 2x$

But, the volume of the compartment must be 8 cubic meters.

⇒ $x^3 + x^2 - 2x = 8$

⇒ $x^3 + x^2 - 2x - 8=0$

⇒ $(x-2)(x^2+3x+4)=0$

If $x-2=0\implies x = 2$ and if $x^2+3x+4=0\implies x = \text{complex number}$

But, width can not be the complex number.

Therefore, width of the compartment = 2 meter.

Length of the compartment = 2 + 2 = 4 meter.

And, Depth of the compartment = 2 - 1 = 1 meter.

Since, the function that shows the volume of the compartment is,

$V(x) = x^3 + x^2 - 2x$

When we lot the graph of that function we found,

V(x) is maximum for infinite.

But width can not infinite,

Therefore, the maximum value of V(x) will be 8.

8 0

3 years ago

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baherus [9]

Answer:

The equivalent ratios are B and C.

4 0

3 years ago

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