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3 years ago

11

Find the inverse for the linear function g(x)=1/4-3

1 answer:

Dimas [21]3 years ago

7 0

G (x) = -11/4 ...

that's your answer

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Select all that apply.

MrMuchimi

Answer:

sorry I don't understand

Step-by-step explanation:

6 0

2 years ago

Please help! i dont understand

kvv77 [185]

QUESTION 1

The given inequality is

$y\leq x-3$ and $y\geq -x-2$ .

If (3,-2) is a solution; then it must satisfy both inequalities.

We put x=3 and y=-2 in to both inequalities.

$-2\leq 3-3$ and $-2\geq -3-2$ .

$-2\leq 0$ :True and $-2\geq -5$ :True

Both inequalities are satisfied, hence (3,-2) is a solution to the given system of inequality.

QUESTION 2

The given inequality is

$y\:>\:-3x+3$ and $y\:>\: x+2$ .

If (1,4) is a solution; then it must satisfy both inequalities.

We put x=1 and y=4 in to both inequalities.

$4\:>\:-3(1)+3$ and $4\:>\: 1+2$ .

$4\:>\:0$ :True and $4\:>\: 3$ :True

Both inequalities are satisfied, hence (1,4) is a solution to the given system of inequality.

Ans: True

QUESTION 3

The given inequality is

$y\leq 3x-6$ and $y\:>\: -4x+2$ .

If (0,-2) is a solution; then it must satisfy both inequalities.

We put x=0 and y=-2 in to both inequalities.

$-2\leq 3(0)-6$ and $-2\:>\: -4(0)+2$ .

$-2\leq -6$ :False and $-2\:>\:2$ :False

Both inequalities are not satisfied, hence (0,-2) is a solution to the given system of inequality.

Ans:False

QUESTION 4

The given inequality is

$2x-y\:$ and $x+y\:>\:-1$ .

If (0,3) is a solution; then it must satisfy both inequalities.

We put x=0 and y=3 in to both inequalities.

$2(0)-3\:$ and $0+3\:>\:-1$ .

$-3\:$ :True and $3\:>\:-1$ : True

Both inequalities are satisfied, hence (0,3) is a solution to the given system of inequality.

Ans:True

QUESTION 5

The given system of inequality is

$y\:>\:2x-3$ and $y\:$ .

If (-3,0) is a solution; then it must satisfy both inequalities.

We put x=-3 and y=0 in to both inequalities.

$0\:>\:2(-3)-3$ and $0\:$ .

$0\:>\:-9$ ;True and $0\:$ :True

Both inequalities are satisfied, hence (-3,0) is a solution to the given system of inequality.

Ans:True

3 0

3 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

<em>Hint: sin is only positive in Quadrants I and II</em>

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

3 years ago

From the chapter Rational Numberssome body please answer fast

Alexxx [7]

Answer:

- $\frac{3}{7}$

Step-by-step explanation:

Given

- $\frac{42}{98}$ ← divide numerator/ denominator by 14 )

= - $\frac{3}{7}$

7 0

2 years ago

Identify the type of conic section that has the equation 4x^2+25y^2=100

Katena32 [7]

The equation you provided is an ellipse; when graphed, it will make an oval shape.

8 0

3 years ago

Read 2 more answers