All categories

3 years ago

how many more pairs of parallel sides does a regular octagon have than a regular octagon have than a regular hexagon

Mathematics

1 answer:

lana [24]3 years ago

5 0

A regular hexagon has 3 pairs of parallel sides. A regular octagon has 4 pairs of parallel sides. Hope this helps :))

You might be interested in

Complete each congruence statement by naming the corresponding angle or side (odd questions only)

Oksi-84 [34.3K]

Given:

1. $\Delta HJI\cong \Delta IQP$

3. $\Delta JLK\cong \Delta FGH$

To find:

The missing values to complete the congruence statements:

1. $\overline{IH}\cong \_\_\_\_$

3. $\angle KJL\cong \_\_\_\_\_$

Solution:

We have,

$\Delta HJI\cong \Delta IQP$

Here, vertices H, J, I are corresponding to I, Q, P. So,

$\overline{IH}\cong \overline{PI}$

Therefore, the complete statement is $\overline{IH}\cong \overline{PI}$ .

We have,

$\Delta JLK\cong \Delta FGH$

Here, vertices J, L, K are corresponding to F, G, H. So,

$\angle KJL\cong \angle HFG$

Therefore, the complete statement is $\angle KJL\cong \angle HFG$ .

4 0

3 years ago

Ou can draw a quadrilateral with two sets of parallel lines and no right angles.

Triss [41]

I believe it would be true because you can have the first set then when you have your second set it just have to be slanted for them to all be acute & obtuse angles

4 0

3 years ago

Please check for me and if any are incorrect please explain!

Anika [276]

Answer: (1) 14.4% (2) 3.78% (3) $222.48 (4) 21,176.47

(5) $425 (6) 14.3 cents per mile

Step-by-step explanation:

1) This is a calculator question. I = 0.144 --> 14.4%

$A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\bullet \text{A = Accrued amount (total amount paid)}\\\bullet \text{P = Principal (initial cost of the car)}\\\bullet \text{r = rate (interest rate in decimal form)}\\\bullet \text{n = number of times in a year (number of months)}\\\bullet \text{t = number of years}\\\\A = m \times nt\\\bullet \text{m = monthly payment amount}\\\bullet \text{nt = number of payments made}\\\implies m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

2) m = 93.33

nt = 36

P = 3000

n = 12

$m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\93.33\times 36=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\3359.88=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.11996=\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.00315198=1+\dfrac{r}{12}\\\\\\0.00315198=\dfrac{r}{12}\\\\\\0.0378=r\\\\\\\large\boxed{3.78\%}=r$

3) same equation as #2 but deduct the down payment from the Principal

nt = 42

P - d = 5,555 - 555 = 5,000

r = 18% --> 0.18

n = 12

$m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\m\times 42=(5,555-555)\bigg(1+\dfrac{.18}{12}\bigg)^{42}\\\\\\42m=5000(1.015)^{42}\\\\\\42m=5000(1.868847)\\\\\\42m=9344.23557\\\\\\m=\large\boxed{222.48}$

4) Gas + Oil = cost per mile × number of miles

1000 + 800 = 0.085x

1800 = 0.085x

$\large\boxed{21,176.47}=x$

5) $\text{Annual Depreciation}=\dfrac{\text{Cost of car - Trade-in value}}{\text{Years driven}}$

$.\qquad =\dfrac{3800-1250}{6}\\\\\\.\qquad =\dfrac{2550}{6}\\\\\\.\qquad = \large\boxed{425}$

6) $\dfrac{\text{(Cost of car - Market value) + Gas & Oil + Parts, Maintenance + Insurance}}{\text{Miles driven}}$

$=\dfrac{(5000-3800)+750+250+300}{17,500}\\\\\\=\dfrac{2500}{17,500}\\\\\\=0.142857\\\\=\large\boxed{14.3\ cents\ per\ mile}$

4 0

3 years ago

Need help explain to me

Natali [406]

9514 1404 393

Answer:

A) (20 miles)/(1 gallon) = 20 miles per gallon

B) 20/1 = 20

Step-by-step explanation:

After you read the problem statement and identify the questions being asked, it is a good idea to look carefully at the given information. Graphs A and B are identical in every way.

The vertical axis is labeled "y" and "Distance (miles)". The horizontal axis is labeled "x" and "Gallons". The graph starts at (0 gallons, 0 miles) and goes through points on the grid that are 2 squares up for each 1 square to the right. Each square up represents 10 miles. Each square to the right represents 1 gallon.

A) Pick two points on the graph. It is usually convenient to choose points where the graph crosses grid intersections. It is often convenient to choose one of them as (0, 0) if that point is on the graph (it is). I expect it to be convenient to choose the second point as (1 gallon, 20 miles).

The change in distance between the point (0 gallons, 0 miles) and (1 gallon, 20 miles) is (20 miles) - (0 miles) = 20 miles.

The change in gallons between the point (0 gallons, 0 miles) and (1 gallon, 20 miles) is (1 gallon) - (0 gallons) = 1 gallon.

The ratio of these changes is (20 miles)/(1 gallon) = 20 miles per gallon.

B) We can use the same two points. Here, we basically ignore the units.

Change in y = 20-0 = 20.

Change in x = 1 -0 = 1.

Slope = 20/1 = 20.

_____

Additional comment

For many problems, I find it convenient to keep the units with the numbers. Making sure the units work out properly is called "units analysis." Doing that can save you from a number of mistakes where units are involved.

Units can be treated just like a variable--multiplied, divided, raised to a power, cancelled from fractions. And, like with variables, terms can only be added or subtracted if they have the same units.

Here, we have ...

$\dfrac{\text{change in distance}}{\text{change in gallons}}=\dfrac{20\text{ miles}}{1\text{ gallon}}=\dfrac{20}{1}\times\dfrac{\text{miles}}{\text{gallon}}=20\text{ miles per gallon}$

Note that "divided by" and "per" mean essentially the same thing in this context.

As you can see from the work done with Graph A, the slope of the line has units that are the ratio of the units of the y-axis to the units of the x-axis. (miles/gallon) Even though in Graph B, we write the slope as 20 without any units, it will only make any sense to use that number where units of miles/gallon make sense.

7 0

3 years ago

What is the slope of this ?

monitta

(1,3)

Hope this helps!
~C

6 0

3 years ago