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Gennadij [26K]
2 years ago
11

Ellis is going on a tour that costs $65 while on vacation. He was told to bring $118 to pay for the cost of the tour and dinner.

Which of the following inequalities represents the money Ellis can spend on dinner?
A. $118 + d < $65
B. $118 + d > $65
C. $65 + d > $118
D. $65 + d < $118
Mathematics
1 answer:
Olegator [25]2 years ago
3 0

Answer:

The answer should be D.

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The stock market gained 45 points on Tuesday and lost 32 points on Wednesday. It had closed on Monday at 2760 points. Where did
Svetllana [295]

Answer:

The market closed with a rating of 2773 points on Wednesday.

Step-by-step explanation:

Lets start with what we know:

Monday at close it had a rating of 2760 points.

Tuesday it gained 45 points.

Wednesday it had lost 32 points.

Once written out it becomes less intimidating to answer.

Monday + Tuesday + Wednesday = Answer

2760 + 45 - 32

= 2805 - 32

= 2773

7 0
3 years ago
Read 2 more answers
What is the equation of a line that passes through (4, 7) and has a slope of −14?
deff fn [24]
Slope = -1/4
(4, 7)

y - 7 = -1/4 (x - 4)
y - 7 = -1/4x + 1
y = -1/4 + 8

answer is C
5 0
3 years ago
Translate this phrase into an algebraic expression:
krek1111 [17]
Jesus is the lord above all
3 0
3 years ago
general admission to the fair is 10.00. each ride costs 4.75. if lisa has 75.00 to spend at the fair, what method can be used to
Elis [28]

Answer:

well the answer is 13

Step-by-step explanation:

The way you can figure this out is subtract 10 from 75 to get 65 and then you divide 65 by 4.75 to get 13 with a huge decimal but you just have 13 because you cant go for part of a ride.

PLEASE PUT BRAINLIEST.

6 0
3 years ago
Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s
WINSTONCH [101]

Answer:  \bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}

<u>Step-by-step explanation:</u>

There are 3 conditions that must be satisfied:

  1. f(x) is continuous on the given interval
  2. f(x) is differentiable
  3. f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2     [0, 3]

1. There are no restrictions on x so the function is continuous \checkmark

2. f'(x) = 3x² - 2x - 6 so the function is differentiable \checkmark

3. f(0) =  0³ - 0² - 6(0) + 2 = 2

   f(3) =  3³ - 3² - 6(3) + 2  = 2

   f(0) = f(3) \checkmark

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1

Only one of these values (1.8) is between 0 and 3.

5 0
3 years ago
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