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Juli2301 [7.4K]

3 years ago

11

List at least three shapes, other than hexagons, in the table. If the shape is a polygon, indicate whether the shape is regular

or irregular. Find the number of lines of reflection that will map each shape back on to itself when the shape flips about the reflection line. Also describe the position of the line of reflection in your own words. Use GeoGebra to guide you in this exercise, if you wish.

2 answers:

aleksklad [387]3 years ago

8 0

Answer:

equilateral triangle 3 3 perpendicular bisectors of the sides

regular heptagon 7 7 perpendicular bisectors of the sides

regular octagon 8 4 perpendicular bisectors and 4 lines joining pairs of opposite vertices

Step-by-step explanation: Plato answer

pentagon [3]3 years ago

3 0

Answer:

equilateral triangle 3 3 perpendicular bisectors of the sides

regular heptagon 7 7 perpendicular bisectors of the sides

regular octagon 8 4 (perpendicular bisectors and 4 lines joining pairs of opposite vertices)

Step-by-step explanation:I got you

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Solve the equation. StartFraction dx Over dt EndFraction equals StartFraction 1 Over x e Superscript t plus 7 x EndFraction An i

MAVERICK [17]

Answer:

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

Step-by-step explanation:

We are given that

$\frac{dx}{dt}=\frac{1}{xe^{t+7x}}$

We have to find the implicit function

Using separation variable method

$\frac{dx}{dt}=\frac{1}{xe^t\cdot e^{7x}}$

By using property $x^a\cdot x^y=x^{a+y}$

$xe^{7x}dx=e^{-t}dt$

By using property $\frac{1}{x^a}=x^{-a}$

Taking integration on both sides

$\int xe^{7x}dx=\int e^{-t}dt$

Parts integration method

$\int u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

By parts integration method

$x\int e^{7x}dx-\int (\frac{dx}{dx}\int e^{7x}dx)dx=-e^{-t}+C$

Using formula $\int e^{ax} dx=\frac{e^{ax}}{a}+C$

$\frac{xe^{7x}}{7}-\frac{1}{7}\int e^{7x}dx=-e^{-t}+C$

$\frac{xe^{7x}}{7}-\frac{1}{49}e^{7x}+e^{-t}=C$

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

We are given that

$F(x,t)=C$

$F(x,t)=\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

8 0

3 years ago

If Triangle ABC is rotated 90 degrees clockwise about the origin, what will be the new coordinates of vertex B?

quester [9]

Answer:

ithe answer to your question is going to be c

6 0

3 years ago

Hi can anyone help me with this please

Nostrana [21]

Answer:35in.

Step-by-step explanation:

I reccomend splitting it into sections when given problems like this.

The smaller two rectangles would be 5in * 2in which is 10in then just add them together.

For the bigger one, you'd do 5-2 in order to get the missing inches which will be 3+2 once added 5in*3in is 15in.

Finally, add 15+10+10 (15=The middle rectangle and 10=the sides)

8 0

3 years ago

Read 2 more answers

use the venn diagram in Part A to find each probability when annie randomly chooses one member of the doubles club. (need help w

Usimov [2.4K]

If we assume the club membership includes

$\bullet\ \mbox{6 junior boys}\\ \bullet\ \mbox{8 junior girls}\\ \bullet\ \mbox{16 senior boys}\\ \bullet\ \mbox{12 senior girls}$

then the blanks in the Juniors circle must be filled in with 6 and 8 (with the 8 going in the overlap with girls). The remaining blank in the Girls circle must be filled in with 12, the number of senior girls. The blank outside both circles is the number of senior boys (members who are not juniors and not girls), 16.

3 0

3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago