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Mekhanik [1.2K]
4 years ago
6

Complete the square to determine the maximum or minimum value of the function defined by the expression. −x2 − 10x + 14 A) minim

um value at 25 B) maximum value at 39 C) maximum value at −5 D) minimum value at −39
Mathematics
2 answers:
Masja [62]4 years ago
4 0
B)maximum value at 39
Gre4nikov [31]4 years ago
4 0

Answer:

B) maximum value at 39.

Step-by-step explanation:

maximum value at 39

First, set the expression equal to 0.

−x2 − 10x + 14 = 0

Complete the square.

−x2 − 10x = −14

−(x2 + 10x) = −14

−(x2 + 10x + 25) = −39

−(x + 5)2 + 39= 0.

Therefore, the maximum value is 39.

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Read 2 more answers
Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim
Ber [7]

The given function is

f(x)=4\cos \text{(}\pi x)+1

The general form of the cosine function is

y=a\cos (bx+c)+d

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}

The equation of the midline is

y_{ml}=\frac{y_{\max }+y_{\min }}{2}

Since the maximum is at the greatest value of cos, which is 1, then

\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}

Since the minimum is at the smallest value of cos, which is -1, then

\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}

Then substitute them in the equation of the midline

\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

3 0
1 year ago
9/15 + 7/15= ?<br><br><br><br><br> Pls help I’m being dumb :0
Mashutka [201]
I think the answer is 1.06666666667
7 0
3 years ago
line m contains the points -3,4 and 1,0. write the equation of a line that would be perpendicular to this one and pass through t
castortr0y [4]

The equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6) in point slope form is y = x + 8

<h3><u>Solution:</u></h3>

Given that line m contains points (-3, 4) and (1, 0)

We are asked to find the equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6)

<em><u>Let us first find slope of the line "m"</u></em>

Given two points are (-3, 4) and (1, 0)

m = \frac{y_2 - y_1}{x_2 - x_1}

\left(x_{1}, y_{1}\right)=(-3,4) \text { and }\left(x_{2}, y_{2}\right)=(1,0)

m=\frac{0-4}{1-(-3)}=-1

Thus slope of line m is -1

We know that <em>product of slope of given line and perpendicular line are always -1</em>

So, we get

\begin{array}{l}{\text { slope of line } m \times \text { slope of perpendicular line }=-1} \\\\ {-1 \times \text { slope of perpendicular line }=-1} \\\\ {\text { slope of perpendicular line }=1}\end{array}

So we have got the slope of perpendicular line is 1 and it passes through (-2, 6)

Let us use the point slope form to find the required equation

<em><u>The point slope form is given as:</u></em>

y - y_1 = m(x - x_1)

(x_1, y_1) = (-2, 6) and m = 1

y - 6 = 1(x - (-2))

y - 6 = x + 2

y = x + 8

Thus equation of required line in point slope form is y = x + 8

7 0
3 years ago
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