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2 years ago

Simplify this question please im really confused

Mathematics

1 answer:

Gemiola [76]2 years ago

8 0

Answer:

option A

Step-by-step explanation:

$\frac{x^a}{x^b} = x^{a-b}$ , $z^0 = 1$

So,

$\frac{18x^5 y}{6xy^8z^0} = \frac{18}{6} \times x^{5 -1} \times y^{1-8} = 3 x^4 \times y^{-7} = \frac{3x^4}{y^7}$

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Which scale factors produce an expansion under a dilation of the original image?

alina1380 [7]

Answer: The correct options are (A) 8 and (D) 7.

Step-by-step explanation: We are given to select the correct scale factors that produces an expansion under a dilation of the original image.

The SCALE FACTOR of a dilation is defined as follows:

$\textup{Scale factor}=\dfrac{\textup{length of a side of the dilated image}}{\textup{length of the corresponding side of the original image}}.$

The dilation will be an expansion is the scale factor is greater than 1 (> 1),

and it will be a reduction if the scale factor is less than 1 (< 1).

Option (A) is

Scale factor = 8.

Since 8 > 1, so the scale factor of the dilation is greater than 1. This implies that the dilation is an expansion.

Option (B) is

Scale factor = 0.7.

Since 0.7 < 1, so the scale factor of the dilation is less than 1. This implies that the dilation is a reduction.

Option (C) is

Scale factor = 0.8.

Since 0.8 < 1, so the scale factor of the dilation is less than 1. This implies that the dilation is a reduction.

Option (D) is

Scale factor = 7.

Since 8 > 1, so the scale factor of the dilation is greater than 1. This implies that the dilation is an expansion.

Thus, the correct options are (A) 8 and (D) 7.

7 0

3 years ago

Read 2 more answers

∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1

igor_vitrenko [27]

It looks like you want to compute the double integral

$\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy$

over the region D with the unit circle x ² + y ² = 1 as its boundary.

Convert to polar coordinates, in which D is given by the set

D = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π}

and

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

Then the integral is

$\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}$