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Vsevolod [243]
3 years ago
13

Aro diagrams his river rafting trip, estimating the time it will take him to paddle upstream against the current, and then back

downstream with the current. His campsite destination is 5.2 miles upstream. Determine how fast Aro can paddle and how fast the river water is moving. Round to the nearest hundredth as needed. Upstream: Downstream: Let x be the speed of Aro’s paddling and let y be the speed of the river. Upstream: 1.04 = x – y Downstream: 2.08 = x + y Aro can paddle at a speed of miles per hour. The river’s speed is miles per hour.
Mathematics
2 answers:
eduard3 years ago
4 0

Answer:

Upstream

5.2=(x-y)5

Downstream

5.2=(x+y)2.5

Step-by-step explanation:

gavmur [86]3 years ago
4 0

Answer: Aro can paddle at a speed of 1.56 miles per hour. The river’s speed is 0.52 miles per hour.

hope this helps (:

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Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
3 0
3 years ago
A recent study of the relationship between social activity and education for a sample of corporate executives showed the followi
jekas [21]

Answer:

Chisquare statistic

Step-by-step explanation:

The most appropriate test to use here is the Chisquare test as it isemployed when testing the relationship between two variables. Both the T statistic and z statistics are the used for testing difference in means. However, in the analysis above, we are given two variables with each having its own levels. Hence, the Chisquare test is the most appropriate in this kind of situation a it measures if a difference exists between the two variables. It tests if they are related or independent of on one another

7 0
3 years ago
The average of six numbers is 4. If the of two of those numbers is 2, what is the average of the other four numbers?
matrenka [14]

Answer:

5

Step-by-step explanation:

total sum of the 6 numbers

= 4 ×6

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= 24 - 2(2)

= 20

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= 20 ÷ 4

= 5

6 0
3 years ago
Gary drove to the park at a rate of 50 miles per hour if it took him 2.5 hours to get from his house to the park how far away is
Marina86 [1]

Answer:

The distance of the park from the Gary's house is 125 miles.

Step-by-step explanation:

Speed of Gary = 50 miles/hour

Time taken by Gary to reach the park from his house = 2.5 hours

Now, we know that,

Distance travelled = speed × time

So,  distance between park and Gary's house = speed × time

= 50 × 2.5

= 125 miles

So, the park is 125 miles away from the Gary's house.

4 0
3 years ago
(02.04 MC)
borishaifa [10]

Answer:

umm idk let me get back to you

Step-by-step explanation:

3 0
3 years ago
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