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3 years ago

An image of a dilation about the origin is shown. Which represents the pre-image of A'?

Mathematics

2 answers:

olasank [31]3 years ago

6 0

Answer:

the answer is D (5/3,-5/3)

Step-by-step explanation:

dimulka [17.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

i just took the test lol

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BartSMP [9]

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3 years ago

On the basis of data from 1990 to 2006, the median income y in the year x for men and women is approximated by the equations giv

worty [1.4K]

Answer:

2080

Step-by-step explanation:

The solution to these equations might best be found using Cramer's Rule. It tells you ...

x = (56939×3 -41655×2)/(-236×3 +838×2) = 87507/968 ≈ 90.400

This is years after 1990, so the year the incomes become the same is predicted to be ...

1990 +90.4 = 2080.4

The median incomes will be the same in the year 2080.

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4 years ago

Need help it’s late and forgot I had homework

Mekhanik [1.2K]

15^ is equivalent to 0.15, or 15/100. If you divide both numerator (15) and denominator (100) by 5, you get 3/20.

28% corresponds to 0.28. What fraction would that be?

Come on. Show your stuff.

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4 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and asymptotes

$y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x$

b) The length of the transverse axis is given by $2a$ . Therefore, the lenght is 6.

c) See below.

4 0

4 years ago