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3 years ago

Joel biked 9 miles east and then 12 miles north. If he

Mathematics

2 answers:

Bumek [7]3 years ago

8 0

Answer:

Step-by-step explanation:

andrey2020 [161]3 years ago

5 0

Answer:

36 miles

Step-by-step explanation:

His entire journey, represented with a diagram, is in the shape of a right angle triangle.

We need to first find the hypotenuse of the triangle, then, add all the sides together:

hyp² = opp² + adj²

hyp² = 9² + 12²

hyp² = 81 + 144 = 225

Find the square root of both sides:

hyp = 15 miles

Therefore, adding the three sides together:

Total distance traveled = 9 + 12 + 15

Total distance traveled = 36 miles

He would ride 36 miles altogether.

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Simplify the expression-2(-9g+4)

xenn [34]

Answer:

18g - 8

Step-by-step explanation:

Multiply using the distributive property.

8 0

3 years ago

Read 2 more answers

Allan plays a video game in which he starts with 0 points. In round 1, he loses 3 1/2 points, in rounds 2, he wins 28 1/2 points

weqwewe [10]

Answer:

Step-by-step explanation:

start with 0

round 1 : lose 3 1/2

round 2: wins 28 1/2

round 3 : lose 3 1/2

0 - 3 1/2 + 28 1/2 - 3 1/2 =

28 1/2 - 7 =

21 1/2 <==== thats 21 and 1/2 points

5 0

3 years ago

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 ...

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

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 ...

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

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

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

_____

<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

Tell whether the lines through the given points are parallel, perpendicular, or neither.

ira [324]

Answer: Line 1: (2, 3) , (4, 12)

m = (12 - 3)/(4 - 2) = 9/2 This is the slope of the line

y = (9/2)x + b

3 = (9/2)(2) + b

3 = 9 + b

b = -6

y = (9/2)x - 6

Line 2: (5, 10) , (14,8)

m = (8 - 10)/(14 - 5) = -2/9

this slope is the opposite sign, and inverse of the first equation's slope. Therefore, the line is perpendicular.

y = mx + b

8 = (-2/9)(14) + b

8 = -28/9 + b

11.11 = b

y = (-2/9)x + 11.11

Step-by-step explanation: Hope this helps :)

7 0

2 years ago

To rent a certain meeting room, a college charges a reservation fee of $18 and an additional fee $5 of per hour. The chemistry c

julsineya [31]

Answer:

You need the minimum cost for renting the meeting room given the following info

Reservation Fee = $18

Hourly Rate = $5

The renter wants to keep the cost below $58

So the cost will vary with the time, t you use the inequality below

5t + 18 ≤ 58

You can solve this to get that minimum

You can check you answer by graphing it.

I hope this helps!<3

3 0

2 years ago