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Sergeeva-Olga [200]
2 years ago
10

Which graph represents the function f (x) = StartFraction 3 x minus 2 Over x minus 2 EndFraction? On a coordinate plane, a hyper

bola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 2, and the horizontal asymptote is at y = 0. On a coordinate plane, a hyperbola is shown. One curve opens up and to the left in quadrant 2, and the other curve opens down and to the left in quadrant 4. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 1. On a coordinate plane, a hyperbola is shown. One curve opens up and to the left in quadrant 2, and the other curve opens down and to the left in quadrant 4. A vertical asymptote is at x = negative 2, and the horizontal asymptote is at y = 4. On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 2, and the horizontal asymptote is at y = 3.
Mathematics
2 answers:
Gennadij [26K]2 years ago
5 0

Answer:

D

Step-by-step explanation:

right on edge

ziro4ka [17]2 years ago
5 0

Last graph on edgeenitu trash website. we all are set up to fail lol

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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
2 years ago
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Darya [45]

The solution to this system is (x, y) = (8, -22).

The y-values get closer together by 2 units for each 2-unit increase in x. The difference at x=2 is 6, so we expect the difference in y-values to be zero when we increase x by 6 (from 2 to 8).

You can extend each table after the same pattern.

In table 1, x-values increase by 2 and y-values decrease by 8.

In table 2, x-values increase by 2 and y-values decrease by 6.

The attachment shows the tables extended to x=10. We note that the y-values are the same (-22) for x=8 (as we predicted above). That means the solution is ...

... (x, y) = (8, -22)

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vesna_86 [32]

Answer:

x=3 meters

Step-by-step explanation:

step 1

Find the area of the rectangular pool

A=LW

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L=18\ m\\W=12\ m

substitute

A=18(12)=216\ m^2

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

L=(18+2x)\ m\\W=(12+2x)\ m

substitute

A=(18+2x)(12+2x)

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

so

A=(18+2x)(12+2x)-216

step 4

Find the value of x if the area of the walkway equal the area of the pool

so

(18+2x)(12+2x)-216=216

Solve for x

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Solve the quadratic equation by graphing

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see the attached figure

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Answer:

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Step-by-step explanation:

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Answer:

8 hours

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57 + 9h = 129

9h = 72

h = 8

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2 years ago
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