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Simora [160]
2 years ago
11

Select all the sequences of reflections that produce an image equivalent to the image r(180°, O)(△BCD).

Mathematics
1 answer:
Aneli [31]2 years ago
8 0

Answer:

r(180°,0) is a rotation of 180° degrees over the origin.

Notice that this rotation moves our figure to the opposite quadrant (so a translation of two quadrants).

Then this is equivalent to:

A reflection over the x-axis followed by a reflection over the y-axis.

Or.

A reflection over the y-axis followed by a reflection over the x-axis.

There is another possible reflection, but it depends on where is our figure.

If the figure is in the first or third quadrant, a reflection over the line y = -x is equivalent to the rotation.

If the figure is in the second or third quadrant, then the reflection over the line y = x is equivalent to the rotation.

We can combine those two and write:

A reflection over the line y = (-1)^n*x.

Where n is the number associated with the quadrant where the figure is in.

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A square with side lengths of 3x + 3. An equilateral triangle with side lengths of 5 x + 0.5. The perimeters of the square and t
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Answer:

1) 4(3x+3)=3(5x+0.5)

2) 3.5

3) 54 units

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2 years ago
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· Andrew purchased some drinks and some chips.
iren [92.7K]

Answer:

C The total amount spent n drinks.

Step-by-step explanation:

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Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat
yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

P(t) = 0.8t^{2} + 6t + 19000

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

P(t) = 0.8t^{2} + 6t + 19000

0.8t^{2} + 6t + 19000 = 34200

0.8t^{2} + 6t - 15200 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

0.8t^{2} + 6t - 15200 = 0

So a = 0.8, b = 6, c = -15200

Then

\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356

t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14

t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0
2 years ago
Anna is at the movie theater and has $35 to spend. She spends $9.50 on a ticket and wants to buy some snacks. Each snack costs $
EastWind [94]

Answer with explanation:

Amount of money Possessed by Anna = $ 35

Money spent on ticket = $9.50

Money spent on Snacks = $ 3.50

Let x number of snacks, which will be least number of snacks that Anna can buy.

transforming the situation in terms of inequality

→9.50 +3.50 x≤ 35

→9.50 -9.50+3.50 x≤35-9.50

→3.50 x≤25.50

Dividing both sides by 3.50, we get

→x≤7.3(approx)

which can't be number of Snacks, as it will be an integral value.

So, minimum number of snacks with given amount of money = 7

So, Anna can buy snacks(x)={x:x≤7,x=1,2,3,4,5,6,7}=At most 7.

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What integer represents a rise in temperature of 19°?
nataly862011 [7]

You would write +19 to indicate an increase of 19

The opposite of that would be -19, read out as negative 19.

7 0
2 years ago
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