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Tanya [424]
2 years ago
7

A trail around a river is 3 1/4 miles long. What is the length, in feet, of the trail?

Mathematics
1 answer:
bezimeni [28]2 years ago
4 0

Answer: 17160 feet

Step-by-step explanation:

We know that 5280 feet = 1 mile

So all we have to do is multiply 3 1/4 by 5280

Let's split the 3 and 1/4 up to make the multiplication easier

3×5280=15840

\frac{1}{4}×5280=\frac{5280}{4} =1320

Now add them together

15840+1320=17160

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After sailing 12 mi, a sailor changed direction and increased the boat's speed by 4 mph. an additional 18 mi was sailed at the i
inessss [21]
Let the speed for the first 12 mi be x mi/h, the speed for 18 mi was (x+4) mi/h
thus given that
time=distance/speed 
the average time will be:
3=(12+18)/(x+x+4)
3=30/(2x+4)
solving for x we get
3(2x+4)=30
6x+6=30
6x=24
x=4 mi/hr
Answer: 12 mi/hr
8 0
3 years ago
A shopkeeper bought some eggs at ¢15 each. Six of them were broken while the rest were sold at ¢20 each. If he made a profit of
zvonat [6]
In the given problem it is already stated that the cost of each eggs is 15 cent. Out the total eggs bought 6 were broken. The rest of the eggs were sold for 20 cents each and the profit made was $4.80.
Cost of the 6 broken eggs = (6 * 0.15)
                                           = 0.9 cents
Let us assume that the number of eggs bought = x
Then we can write the equation as
0.2x - (0.15x - 0.9) = 4.80
0.2x - 0.15x + 0.9 = 4.80
0.05x = 4.80 - 0.9
0.05x = 3.9
x = 3.9/0.05
   = 78
So the number of eggs bought by him is 78
6 0
3 years ago
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
Whitepunk [10]

If f(x)=4x-3:

\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4

If f(x)=4x^{-3}:

\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}

\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}

\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}

7 0
3 years ago
Let f(x) = ............
34kurt

Answer:

So first lets find g(-1)

so we plug in the answers:

-2 * -1 ^2 -4

-2*1-4 = -2-4 =

-6

Now lets solve for f(-2)

-2^2-3

4-3=1

-6+1 = 5

3*5 = 15

Im getting answer of 15 but it shows -15 or something, so I dont know if I got it wrong or if its like a dash then the answer.

3 0
3 years ago
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