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3 years ago

HELP ME!! WILL MARK BRAINLIEST DUE TODAY!! there are two parts to it !

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

4 0

Answer:

Part A: A

Step-by-step explanation:

Part B: -6 - x - 0

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caleb wants to rent a kayak. Kayak rental cost 14.50 for a half hour.If caleb rents a kayak for one hour and forty-five minutes.

jolli1 [7]

50.75
you multiply 14.50 times 3.5
the reason why it is 3.5 is because there are two half hours in one hour, and another half hour plus 15 minutes ( which is half of a half hour)
i hope this is clear!!
good luck!!

6 0

3 years ago

A motorboat can maintain a constant speed of 44 miles per hour relative to the water. The boat makes a trip upstream to a certai

serg [7]

Answer:

Step-by-step explanation:

Distance upstream = (34-c)(39/60) miles

--------------------------

Distance downstream = (34+c)(29/60) miles

-----------------------------------------------

Equation:

distance = rate*time

distance up = distance down

(34-c)(39/60) = (34+c)(29/60)

Multiply both sides by 60 to get:

39(34-c) = 29(34+c)

39*34 - 39c = 29*34 + 29c

68c = 340

c = 5 mph speed of the current

4 0

3 years ago

For the linear equation 3x + 7y = 42: a. Determine the slope: b. Determine y- intercept if it exists: c. Express equation in slo

algol13

Answer: The required answers are

(a) the slope of the given line is $-\dfrac{3}{7}.$

(b) y-intercept exists and is equal to 6.

(c) the slope-intercept form of the line is $y=-\dfrac{3}{7}x+6.$

Step-by-step explanation: We are given the following linear equation in two variables :

$3x+7y=42~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to :

(a) determine the slope,

(b) determine the y-intercept, if exists

and

(c) express equation in slope-intercept form.

We know that

The SLOPE_INTERCEPT form of the equation of a straight line is given by

$y=mx+c,$ where m is the slope and c is the y-intercept of the line.

From equation (i), we have

$3x+7y=42\\\\\Rightarrow 7y=-3x+42\\\\\Rightarrow y=\dfrac{-3x+42}{7}\\\\\\\Rightarrow y=-\dfrac{3}{7}x+6.$

Comparing with the slope-intercept form, we get

$\textup{slope, m}=-\dfrac{3}{7},\\\\\\\textup{y-intercept, c}=6.$

Thus,

(a) the slope of the given line is $-\dfrac{3}{7}.$

(b) y-intercept exists and is equal to 6.

(c) the slope-intercept form of the line is $y=-\dfrac{3}{7}x+6.$

3 0

3 years ago

Complement and supplement of a 19

Tems11 [23]

well first the complement I 90-19=71

and supplement is 180-19=161

3 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

2 years ago