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anygoal [31]
3 years ago
15

Six kosher hot dogs cost $18.60.What would five of these cost?​

Mathematics
2 answers:
KiRa [710]3 years ago
8 0
18.6/6=3.1
1 hotdog=3.1
So, 3.1*5=15.50$
timurjin [86]3 years ago
3 0

Answer:

$15.58

Step-by-step explanation:

18.60 divided by 6 = 3.01

18.60 - 3.01 = 15.58

Since it is one less than six. we just need to take away 1 part of the divided amount

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Find the measure of the angles indicated
GalinKa [24]

Both angles have a measure of 125°.

<h3>How to find the measure of the angles?</h3>

Assuming the two horizontal lines are parallel, we know that the measure of the two indicated angles must be the same one.

Then:

-1 + 14x = 12x + 17

Now we need to solve this for x:

14x - 12x = 17 + 1

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x =18/2 = 9

Now that we know the value of x we can replace it in any of the expressions for the angles:

-1 + 14*9 = 125°

Both angles have a measure of 125°.

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brainly.com/question/25716982

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8 0
1 year 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
2 years ago
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