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Sati [7]
4 years ago
13

Select the values for p, q, and r that will make the following equation true.

Mathematics
1 answer:
gregori [183]4 years ago
5 0

Answer:

P= 3

Q= -14

R= -5

Step-by-step explanation:

Just solve the equation on the right side, you'll get:

=3x^2 -15x +x -5

=3x^2 -14x -5

Now compare it to the equation on the left side,

The coefficient of x^2 on the left is "p", on the right that we just solved is 3.

Same for "q" which is the coefficient on the left, on the right it's -14.

And for "r" it's -5 from the equation on the right.

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About 250 attendance

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Brad had a small gathering at a local steakhouse. The steakhouse offers three dinner platters which vary by size and price. They
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Hello!

If Brad had a small gathering at a local steakhouse where they ordered 4 of the 6-ounce platters, 3 of the 8-ounce platters, and 2 of the 11-ounce platters. After adding an 18% gratuity their total before sales tax should have been $130.45

Step-by-step explanation:

4 x 9.95=39.80

3 x 12.95=38.85

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5 0
1 year ago
Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from th
fredd [130]

Using continuity concepts, the answers are given by:

  • The function is right-continuous at x = 0.
  • The function is left-continuous at x = 1.

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

A function f(x) is continuous at x = a if:

\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)

  • If \lim_{x \rightarrow a^{-}} f(x) = f(a), the function is left-continuous.
  • If \lim_{x \rightarrow a^{+}} f(x) = f(a), the function is right-continuous.
  • For the given function, the continuity is tested at the points in which the definitions are changed, x = 0 and x = 1.

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

At x = 0.

  • Approaching from the left, it is less than 0, thus:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} x + 9 = 0 + 9 = 9

  • Approaching from the right, it is more than 0, thus:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} e^x = e^0 = 1

  • The numeric value is:

f(x) = e^0 = 1

  • Since [tex]\lim_{x \rightarrow 0^+} f(x) = f(0), the function is right-continuous at x = 0.

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

At x = 1.

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} e^x = e^1 = e

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} 4 - x = 4 - x = 3

f(1) = e^1 = e

  • Since [tex]\lim_{x \rightarrow 1^-} f(x) = f(1), the function is left-continuous at x = 0.

A similar problem is given at brainly.com/question/21447009

7 0
3 years ago
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