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Natalija [7]
3 years ago
6

An electrician charges a set fee of $50 for every house call and then charges an additional hourly rate. One day, the electricia

n earned a total of $400 for a 5 hour job. Write an equation for C,C, in terms of t,t, representing the total cost of the electrician's services if the electrician spends tt hours at the house working.
Mathematics
1 answer:
Serhud [2]3 years ago
7 0

Answer:

50+5t=C

Step-by-step explanation:

Suppose the electrician spends t hours at the house working.

Amount charged for every house call by electrician = $50

The electrician earned a total of $400 for a 5 hour job.

Also, C denotes total cost of the electrician's services.

So, the equation for C in terms of t is given below.

50+5t=C

Here, C=400

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A. x+x+x+3x+3x=9<br> B. x+x+x+x+x=18<br> C. x+x+x+3x+3x=18<br> D. 9x^5=18
lisov135 [29]

9514 1404 393

Answer:

  a)  C  x +x +x +3x +3x = 18

  b)  x = 2

  c)  2 cm, 6 cm

Step-by-step explanation:

a) The sum of the lengths of the sides is equal to the length of the perimeter. The appropriate equation is ...

 x +x +x +3x +3x = 18

__

b) Simplifying, the equation becomes ...

  9x = 18

  x = 2 . . . . . divide by 9

__

c) The shorter sides are x = 2 cm.

The longer sides are 3x = 6 cm.

6 0
2 years ago
Two airplanes are flying to the same airport. Their positions are shown in the graph. Write a system of linear equations that re
wariber [46]

Answer:

Airplane #1 equation: y=5/13x+42/13

Airplane #2 equation: y=1/3x+14

Step-by-step explanation:

So to find the slope of each airplane, you use the formula y2-y1/x2-x1. That means, for airplane#1 the equation will be 9-4/15-2. Simplify this and get 5/13. Then, for airplane#2, the equation will be 12-9/6-15. Simplify this and get 3/-9 and divide each side by 3 to get 1/-3 or -1/3. Next, use point slope formula to find the system of linear equations. Point slope formula is y-y1=m(x-x1). M is the slope. Use any point from the line. In this case, I will use (2,4). Tat means the first airplane's equation would be y-4=5/13(x-2). Then y-4=5/13x-10/13. Then, convert four into a fraction with a denominator of 13. This means, you have to multiply 4 by 13 to get 52/13. Add 52/13 to -10/13 to get 42/13. That means the first equation will be y=5/13x+42/13. The second equation point will be (6,12). This means the equation will be y-12=-1/3(x-6). Simplify this to get y-12=-1/3x+2. Simplify this to get y=1/3x+14. Therefore, Airplane#1 equation will be y=5/13x+42/13 and airplane #2 equation will be  y=1/3x+14.

Hope this helps

6 0
3 years ago
????????!!!!!!!!!!!!!!????
densk [106]

Answer:

B. 2106 pi mm³

Step-by-step explanation:

the solution is attached,

<h2><u>MARK ME AS BRAINLIST</u> </h2>

8 0
2 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
2 years ago
I need help with this question, I will give brainliest
PolarNik [594]

Answer:

3

Step-by-step explanation:

trust me

5 0
3 years ago
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