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3 years ago

10

The line graphed below has a slope of ____. -3 -1/3 1/3 3

2 answers:

daser333 [38]3 years ago

6 0

-3 because for every 1 you go across you go down 3 and the graph is negative so negative gradient

GalinKa [24]3 years ago

4 0

The slope would be -3

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

2 years ago

Micheal buys a new jacket for 90 dollars. Sales tax is 8%. How much does Micheal actually have to pay?

Illusion [34]

Answer:

$97.20

Step-by-step explanation:

you make the 8% into a decimal by moving the decimal to places to the left.

8%=0.08, Then you multiply that by the whole number of 90 dollars.

0.08 x 90 = $97.20

3 0

2 years ago

Please can you guy help for this questiong

ad-work [718]

Answer:

Reason 1. Given

Statement 2. <CAD is congruent to <BAD

Reason 2. Definition of angle bisector

Statement 3. Segment AD is congruent to segment AD

Reason 4. All right angles are congruent

Reason 5. ASA

4 0

2 years ago

Applications 24. A rocket is launched into the air. Its height in feet, after x seconds, is given by the equation The starting h

SVEN [57.7K]

The given function is

$h(x)=-16x^2+300x+20$

According to this function, the starting height of the rocket is 20 feet because that's the initial condition of the problem stated by the independent term.

Additionally, we find the maximum height by calculating the vertex of the function V(h,k).

$h=-\frac{b}{2a}$

Where a = -16 and b = 300.

$\begin{gathered} h=-\frac{300}{2(-16)}=\frac{300}{32}=\frac{150}{16}=\frac{75}{8} \\ h=9.375 \end{gathered}$

Then, we find k by evaluating the function

$\begin{gathered} k=-16(9.375)^2+300(9.375)+20 \\ k=-1406.25+2812.5+20=1426.25 \end{gathered}$

Hence, the maximum height is 1426.25 feet.

At last, to know the time need to hit the ground, we just use h=9.375 and we multiply it by 2

$t=2\cdot9.375=18.75$

Hence, the rocket hits the ground after 18.75 seconds.

6 0

7 months ago

point T"(-2,5) is a vertex of triangle T"O"P". The original image was rotated 90° counterclockwise around the origin and rhen tr

emmainna [20.7K]

ANSWER

T(-2,1)

EXPLANATION

Let T(a,b) be the coordinates.

When this point is rotated 90° counterclockwise about the origin,

Then,

$T(a,b)\to \: T'( - b,a)$

If this point is then translated using the rule;

$(x,y)\to (x-1,y+7)$

then,

$T(a,b)\to \: T'( - b,a) \to T"( - b - 1,a + 7)$

It was given that, T"(-2,5)

This implies that,

-b-1=-2

-b=-2+1

-b=-1

b=1

a+7=5

a=5-7

a=-2

Therefore the coordinates of T are:

(-2,1)

4 0

3 years ago

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