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Romashka [77]
3 years ago
6

On a particular day, the wind added 5 miles per hour to Jaime's rate when she was rowing with the wind and subtracted 5 miles pe

r hour from her rate on her return trip. Jaime found that in the same amount of time she could row 60 miles with the wind, she could go only 30 miles against the wind.What is her normal rowing speed with no wind?
Mathematics
1 answer:
AleksandrR [38]3 years ago
8 0
Jamie's normal rowing rate with no wind was 60 or 65 miles per hour

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X=67/2
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What is the primary goal and secondary goal of the Fried Liver Opening in chess? <br />
polet [3.4K]

Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.

Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.


3 0
3 years ago
Find an equation of the line with the given slope and​ y-intercept.
vlabodo [156]

Answer:

y=-2x-13

Step-by-step explanation:

y-y1=m(x-x1)

y-(-13)=-2(x-0)

y+13=-2x

y=-2x-13

3 0
2 years ago
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
3 years ago
A group of college students built a self-guided rover and tested it on a plane surface. They programmed the vehicle to move alon
makkiz [27]

Answer:

B. 40 meters

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— Estimating

The rectangle enclosing the path has sides of length 9 m and 14 m, so its perimeter is 2(9+14) = 46 m. The distance covered will be shorter than that.

The distance from A to C is longer than the distance from D to C, so we know the distance will be longer than 2·14+9 = 37 m.

Only one answer choice fits in the range 37 < d < 46.

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— Detailed calculation

The distance from B to C is the hypotenuse of a right triangle with sides 9 and 12. You will recognize that these side lengths are 3 times the side lengths of a 3-4-5 right triangle, so the hypotenuse distance is 3·5 = 15 meters.

The circuit length is ...

AB +BC +CD +DA = 2 + 15 + 14 + 9 = 40 . . . . meters

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3 years ago
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