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Vedmedyk [2.9K]
3 years ago
11

PLEASE HELP ASAP!! WILL GIVE POINTS AND MARK BRAINLIEST!!!!!

Mathematics
1 answer:
Korvikt [17]3 years ago
6 0

meme break

⣠⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠀⠀⠀⠀⣠⣤⣶⣶ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠀⠀⠀⢰⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣀⣀⣾⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⡏⠉⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⣿ ⣿⣿⣿⣿⣿⣿⠀⠀⠀⠈⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠛⠉⠁⠀⣿ ⣿⣿⣿⣿⣿⣿⣧⡀⠀⠀⠀⠀⠙⠿⠿⠿⠻⠿⠿⠟⠿⠛⠉⠀⠀⠀⠀⠀⣸⣿ ⣿⣿⣿⣿⣿⣿⣿⣷⣄⠀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠠⣴⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⠀⢰⣹⡆⠀⠀⠀⠀⠀⠀⣭⣷⠀⠀⠀⠸⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠈⠉⠀⠀⠤⠄⠀⠀⠀⠉⠁⠀⠀⠀⠀⢿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⢾⣿⣷⠀⠀⠀⠀⡠⠤⢄⠀⠀⠀⠠⣿⣿⣷⠀⢸⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⡀⠉⠀⠀⠀⠀⠀⢄⠀⢀⠀⠀⠀⠀⠉⠉⠁⠀⠀⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿

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E<span>xactly one plane contains a given line and a point not on the line.</span>The statement that best describes the relationship between a line and a point in a plane is that, there is exactly one plane containing a given line and a point not on the line. 
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F(x) = 49 − x2 from x = 1 to x = 7; 4 subintervals
Setler [38]

Answe

Given,

f(x) = 49 − x² from x = 1 to x = 7

n = 4

\Delta x = \dfrac{7-1}{4}= 1.5

For x= 1

f(x₀) = 49 - 1^2 = 48

x = 2.5

f(x₁) = 42.75

x = 4

f(x₂) = 49 - 4^2 = 33

x = 5.5

f(x₃) = 49 - 5.5^2 = 18.75

x = 7

f(x₄) = 49 - 7^2 = 0

We have to evaluate the function on therigh hand point

A = \Delta x [f(x_1)+f(x_2)+f(x_3)+f(x_4)]

A = 1.5 [42.75+33+18.75+0]

A = 141.75

For Area for left hand sum

A = \Delta x [f(x_0)+f(x_1)+f(x_2)+f(x_3)]

A = 1.5 [48+42.75+33+18.75]

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3 0
3 years ago
Find the relative minimum of
Harlamova29_29 [7]

Answer:

Step-by-step explanation:

The max and min values exist where the derivative of the function is equal to 0. So we find the derivative:

y'=9x^2+28x-11

Setting this equal to 0 and solving for x gives you the 2 values

x = .352 and -3.464

Now we need to find where the function is increasing and decreasing.  I teach ,my students to make a table.  The interval "starts" at negative infinity and goes up to positive infinity.  So the intervals are

-∞ < x < -3.464           -3.464 < x < .352             .352 < x < ∞

Now choose any value within the interval and evaluate the derivative at that value.  I chose -5 for the first test number, 0 for the second, and 1 for the third.  Evaluating the derivative at -5 gives you a positive number, so the function is increasing from negative infinity to -3.464.  Evaluating the derivative at 0 gives you a negative number, so the function is decreasing from -3.464 to .352.  Evaluating the derivative at 1 gives you a positive number, so the function is increasing from .352 to positive infinity.  That means that there is a min at the x value of .352.  I guess we could round that to the tenths place and use .4 as our x value.  Plug .4 into the function to get the y value at the min point.

f(.4) = -48.0

So the relative min of the function is located at (.4, -48.0)

3 0
3 years ago
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