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jonny [76]
3 years ago
13

Ellen has $75 in her savings account. She deposits $25 every week. Her father also deposits $35 into the account every time Elle

n mows the lawn. Her savings account balance can be shown with the following expression: 75 + 25w + 35m Part A: Identify a coefficient, a variable, and a constant in this expression. (3 points) Part B: If Ellen saves for 8 weeks and mows the lawn 2 times, how much will she have in her account? Show your work to receive full credit. (4 points) Part C: If Ellen had $95 in her savings account, would the coefficient, variable, or constant in the expression change? Why?
Mathematics
2 answers:
Oksanka [162]3 years ago
5 0
-----------------------------------------<span>
A coefficient is the '25' in 25m
A variable is the 'm' in 35m                 <-------- Part A
A constant is the 75.
-----------------------------------------</span><span>
8x25=200
35x2=70                                        <----------- Part B
200+70=270+75=345
Ellen will have $345.
-----------------------------------------
Yes, but only the constant would change. 
The new expression would be: 95+25w+35m.                              <----- Part C
The </span><span>coefficient and variable wouldn't change because she still
deposits $25 a week and gets $35 every time she mows the lawn.</span>
Anna [14]3 years ago
4 0
Saving = $75 + 25.w + 35.m

A. 75 = constant, since it's a deposit not related to any transaction
w and m are 2 variables
25 and 35 are 2 coefficient to w and m respectively

B. Saving = 75 + 25(8) + 35(2)
Saving = $345

C. If Ellen had $95 instead of $75, we will just replace $75 with $98, this will NOT AT ALL AFFECT neither the duration (number of weeks) nor the number of times she mows the loan. Only previously said ONLY The Coefficient will change

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I have corner points of:
WARRIOR [948]
The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10

So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)

Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z

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

Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later. 
So let's call it A. Let A = 25

Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later

Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later

Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727

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

In summary, we found
A = 25
B = 21
C = 39
D = 22.2727

The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)

Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)

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

Final Answer: 39

7 0
3 years ago
Which equation is correctly solved for y​
Radda [10]

Answer:

c

Step-by-step explanation:

5 0
3 years ago
F(x) = x(x - 1)<br><br> g(x) = 3x<br> find (f*g)(6)
svetlana [45]

Step-by-step explanation:

<u>Given functions:</u>

  • f(x) = x(x - 1)
  • g(x) = 3x

<u>Find (f*g)(6):</u>

  • (f*g)(6) = f(6)*g(6) = 6(6 - 1)*3(6) = 30*18 = 540

<u>In case it is a composite function (f · g)(6) the answer is different:</u>

  • (f · g)(6) = f(g(6)) = f(3*6) = f(18) = 18*(18 - 1) = 306
4 0
2 years ago
Read 2 more answers
How would you shift the function
Otrada [13]

I believe that the answer would be:

g(x) = 12x + 17  

5 0
3 years ago
In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

8 0
3 years ago
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