All categories

3 years ago

Dayton is due north of Richmond, and Easton is 16 mi due east of Richmond. Dayton is 32 mi from Easton.

2 answers:

8 0

D
|
|
|
|
R ---------E

RE = 16mi
DE = 32 mi

DR^2 = DE^2 - RE^2 = (32mi)^2 - (16mi)^2 = 768 mi^2

DR = √(768 mi^2) = 27.7 mi

max2010maxim [7]3 years ago

3 0

Given:
Dayton
--32 mi--
Richmond -- 16 mi-- Easton

I can't draw a line but the figure drawn based on the given description is a RIGHT TRIANGLE.

Richmond to Easton is the short leg.
Dayton to Easton is the hypotenuse.
Richmond to Dayton is the long leg.

We are looking for the measure of the long leg. We will use the Pythagorean theorem.

a² + b² = c²
a² + 16² = 32²
a² + 256 = 1,024
a² = 1,025 - 256
a² = 768 mi²
a = √768 mi²
a = 27.71 mi This is the distance of Dayton from Richmond

You might be interested in

Conduct a hypothesis test to test the hypothesis that cloud seeding with silver nitrate increases rainfall. Use significance lev

masya89 [10]

Answer:

Paired t-test

Step-by-step explanation:

paired t–test is used when there is one measurement variable and two nominal variables. The most common design is that one nominal variable represents individual organisms, while the other is "before" and "after" some treatment.

6 0

3 years ago

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

3 0

3 years ago

A season has 81 home games. If you

skelet666 [1.2K]

Answer:

972$

Step-by-step explanation:

8 0

3 years ago

The perimeters of similar triangles are in the same ratio as the corresponding sides

wariber [46]

Answer:

Always

Step-by-step explanation:

Suppose you have triangle ABC with side lengths a, b, c. Suppose that is similar to triangle DEF with side lengths d, e, f.

Now, let k be the ratio of corresponding sides ...

k = d/a

Because the same factor applies to all sides, we also have ...

k = e/b = f/c

That is, if we multiply by the denominators of each of these fractions, we get ...

d = a·k
e =b·k
f = c·k

The perimeter of ΔABC is ...

perimeter(ABC) = a + b + c

The perimeter of ΔDEF is ...

perimeter(DEF) = d + e + f = a·k + b·k + c·k

perimeter(DEF) = k(a + b + c) = k·perimeter(ABC)

k = perimeter(DEF)/perimeter(ABC)

That is, the perimeters are in the same ratio as corresponding sides.

3 0

3 years ago

Blanca runs 8 laps around the track each day to train for an endurance race. She times each lap to practice her pacing for the r

Artist 52 [7]

Answer:

A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds

Step-by-step explanation:

The given table is presented as follows;

$\begin{array}{ccc}Day \ 1 \ lap \ times \ (seconds)&Day \ 2 \ lap \ times \ (seconds)&Day \ 3 \ lap \ times \ (seconds)\\83&87&85\\92&90&86\\91&92&91\\89&91&93\\94&92&91\\93&95&89\\88&90&88\\84&85&84\end{array}$ The number of laps in the range 82 to 84 seconds = 1

The number of laps in the range 84 to 86 seconds = 4

The number of laps in the range 86 to 88 seconds = 2

The number of laps in the range 88 to 90 seconds = 4

The number of laps in the range 90 to 92 seconds = 6

The number of laps in the range 92 to 94 seconds = 5

The number of laps in the range 94 to 96 seconds = 2

The number of laps in the range 96 to 98 seconds = 0

Therefore, the histogram that represents Blanca's lap times for the three days of practice is described as follows;

5 0

3 years ago