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3 years ago

At which x-value does the cosine function attain a maximum value?

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

7 0

9514 1404 393

Answer:

x = 0

Step-by-step explanation:

It can help to keep a graph of the cosine function in mind. It starts off at its maximum value at x = 0.

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HELP ME OUT PLEASE URGENT

gizmo_the_mogwai [7]

I think its 10

Correct me if im wrong

But all u have to do is multiply 2.5 x 4=10

7 0

3 years ago

Number 1d please help me analytical geometry

lesantik [10]

For a) is just the distance formula

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
% (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}$
-----------------------------------------------------------------------------------------
for b) is also the distance formula, just different coordinates and distance

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
% (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}$
--------------------------------------------------------------------------
for c) well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}$

now.. whatever that is, is = BC, so the distance for BC is

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
% (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}$

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d) we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
% (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\
$

$\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}$

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"

7 0

3 years ago

The costprice of the article is Rs800.If the article is sold for Rs1200,find profit

romanna [79]

Answer:

Answer

Profit = SP - CP

Profit = 1200 - 800

Profit = 400

Henceforth,

Profit made on the article is ₹400

8 0

3 years ago

There are two parking garages in beacon falls . Garage a charges $7.00 to park for the first 2 hours ,and each additional hour c

Julli [10]

Answer

Find out the number of hours when the cost of parking at both garages will be the same.

To prove

As given

There are two parking garages in beacon falls .

As given

Let us assume that the y is representing the cost of parking at both garages will be the same.

The total number of hours is represented by the x.

First case

Garage a charges $7.00 to park for the first 2 hours ,and each additional hour costs $3.00 .

As garage charges $7.00 for the first 2 hours so the remaning hours are (x -2)

Than the equation becomes

y = 3.00 (x -2) + 7.00

written in the simple form

y = 3x - 6 +7

y = 3x + 1

Second case

Garage b charges $3.25 per hour to park.

than the equation becomes

y = 3.25x

Compare both the equations

3x +1 = 3.25x

3.25x -3x = 1

.25x = 1

$x = \frac{1}{.25}$

x = 4hours

Therefore in the 4 hours the cost of parking at both garages will be the same.

3 0

3 years ago

A helicopter hovers 500 feet above a small island. The figure shows that the angle of

Tamiku [17]

Answer:

435 ft

Step-by-step explanation:

Reference angle = <P = 49° (angle of elevation)

Opposite side = 500 ft

Adjacent side = d = ?

Apply trigonometric function, TOA.

This:

Tan P = opp/adj

Tan 49 = 500/d

d × Tan 49 = 500

d = 500/Tan 49

d = 434.643369 ≈ 435 ft (to the nearest foot)

6 0

3 years ago