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3 years ago

15

she can go 6 miles upstream in one and a half hours. going downstream, she goes 24 miles in the same time. determine the speed a

t which she kayaks and the speed of the river's current

1 answer:

defon3 years ago

4 0

she is going 4mpr. upstream and 16mpr. downstream

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3 = 2 – x 3 = 2 – 2 – x 3 = –x 2003-03-02-00-00_files/i0110000.jpg –3 = x A. The student incorrectly applied the multiplication

rewona [7]

Answer:
B. The student did not properly apply the addition property to isolate x

Explanation:
When given an equation to solve, always remember that when you do an external operation (add/subtract/multiply a term or divide by a term) on one side of the equation, the same operation should be applied on the other side in order to maintain the equality of the equation.

Now, let's take a look on the steps done:
Step 1:
3 = 2 - x
Step 2:
3 = 2 - 2 - x
Step 3:
3 = -x

Now, note n step 2, the student wanted to get rid of the 2 next to the x, therefore, he subtracted 2. However, the student did not subtract the 2 from the other side of the equation. Since we're taking addition (we're adding a -2), therefore, the student incorrectly applied the addition property to isolate the x.

The correct steps would be as follows:
Step 1:
3 = 2 - x
Step 2:
3 - 2 = 2 - 2 - x
Step 3:
1 = - x

Hope this helps :)

5 0

3 years ago

Pls help fast! Will mark first correct answer as brainliest!!!

hammer [34]

Answer:

the first one is 267% and 400%

Step-by-step explanation:

but i don't know for the 2nd

6 0

3 years ago

Read 2 more answers

Help help please please help hurry ASAP

labwork [276]

1. (2,-1)

2. (1.5,4)

3. (-2,3)

4. (1,4)

5. (0,3)

6. 5,4

7. 0,4

8. -1,-1

3 0

3 years ago

joja [24]

Answer:

Slope of the line x +2y =14 will be -1/2 since y = -1/2x+7, where m = -1/2 and c = 7.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago