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3 years ago

Which of the following is a step used in the construction of an equilateral triangle inscribed in a circle?

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

3 0

It would be better with the image, I think it is the 3 one

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The first term of a geometric sequence is 10,000 and the common ratio is 0.4. What

Misha Larkins [42]

$a = 10000$

$r = 0.4 = 4/10 = 2/5$

$a_9 = ar^8 = 10000 × 2^8/5^8$

$a_9 = 10000 × 256/(25 × 25 × 25 × 25)$

$a_9 = 6.5536$

7 0

2 years ago

maria brought a blouse on sale for 20% off. the sale price was $21.76. what was the original price? round to the nearest hundred

Degger [83]

Answer: $27.20

Step-by-step explanation:

She bought a blouse for 20% off so it means that she paid 80% of the original price.We can then set up the equation

80% of x = 21.76 where x is the original price so solve for x

0.8x = 21.76

x = 27.20

Which means the original price is $27.20

4 0

3 years ago

Logan saved $35 from cutting the neighbor’s yard. He earns $10 a week from washing dishes, which he also saves. Which function c

balandron [24]

the answer is “s = 10w + 35” since he already has 35 but needs to find out how many weeks he washed the dishes :))

7 0

2 years ago

What do you get when you combine the following numbers? -4-4-4-4

sladkih [1.3K]

Answer:

-16

Step-by-step explanation:

you just add all of them together

4 0

2 years ago

Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north

Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

$x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

Child 1 final position => $x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t$
Child 2 final position => $x_{2f} =0+v_{2} t$

4) Sum the equations and equate to 3

$x_{1f} +x_{2f} =3$

5) Substitute the values we already know

$1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]$

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

7 0

3 years ago