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3 years ago

PLEASE HELP ME! I'v tried but still haven't got it!

Mathematics

1 answer:

spayn [35]3 years ago

5 0

Answer:

Step-by-step explanation:

part a: time, student surf the internet

part b

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Which ordered pair is in the solution set of the system of linear inequalities? y >3/2 x – 1 y <3/2 x – 1 (–5, 2) (2, 2) (

Vikentia [17]

Answer:

No solution

Option 4 is correct

Step-by-step explanation:

Given: The system of inequality

$y>\dfrac{3}{2}x-1$

$y$

We need to choose correct option.

Option 1: (-5,2)

Put x=-5 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(-5)-1$ $2$

$2>-8.5$ $2$

TRUE FALSE

Option 2: (2,2)

Put x=2 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(2)-1$ $2$

$2>2$ $2$

FALSE FALSE

Option 3: (5,2)

Put x=-5 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(5)-1$ $2$

$2>6.5$ $2$

FALSE TRUE

Option 1, 2 and 3 are incorrect because both equation must be true for solution.

Hence, Option 4 is correct no solution.

6 0

3 years ago

Read 2 more answers

COS (377= 32

Elanso [62]

The answer is c because the remaining number would be 10

4 0

3 years ago

Determine the inverse of the function f (x) = 4(x − 3)2 + 2. inverse of f of x is equal to 3 minus the square root of the quanti

Elodia [21]

The inverse of the function f(x) = 4(x-3)² + 2 is $f^{-1}(x) = \sqrt{\frac{x-2}{4} } + 3$

The given function is:

f(x) = 4(x - 3)² + 2

To find the inverse of the function:

Make x as the subject of the formula

$4(x-3)^2 = f(x) - 2\\(x-3)^2 = \frac{f(x)-2}{4} \\x - 3 = \sqrt{\frac{f(x)-2}{4} } \\x = \sqrt{\frac{f(x)-2}{4} } + 3$

Replace x by $f^{-1}(x)$ and replace f(x) by x

$f^{-1}(x) = \sqrt{\frac{x-2}{4} } + 3$

Therefore, the inverse of the function is:

$f^{-1}(x) = \sqrt{\frac{x-2}{4} } + 3$

Learn more here: brainly.com/question/17285960

4 0

2 years ago

What is the slope, point, x-intercept, and y-intercept?

IceJOKER [234]

X-int =(-3,0)
Y-int=(0,_1)
Slope: _1/3

5 0

3 years ago

Read 2 more answers

Please help I don’t know this question .

serg [7]

x = 6

AB = 11

BC = 22

Hope that helps :)

6 0

3 years ago