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2 years ago

Which region represents the solution to the given system of inequalities?

Mathematics

1 answer:

agasfer [191]2 years ago

7 0

Answer:

The intersection region shown in the graph attached is the solution of the system of inequalities

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Find the equation of quadratic function determined from the graph below?

Aleksandr [31]

Step-by-step explanation:

The x-intercepts are x = -1 and x = 5, so:

y = k (x + 1) (x − 5)

The vertex is (2, -3), so:

-3 = k (2 + 1) (2 − 5)

-3 = -9k

k = 1/3

y = 1/3 (x + 1) (x − 5)

Simplifying:

y = 1/3 (x² − 4x − 5)

y = 1/3 x² − 4/3 x − 5/3

3 0

3 years ago

What is the value of x? 155 105 A. 55° B. 160° C. 50° O D. 20°

Hatshy [7]

If you are taking one away from the other the answer is 50 degrees so try that hope this helps
=C.

3 0

2 years ago

Leslie decides to join a gym. She must pay a monthly fee plus a one time fee to open a membership. This situation can be modeled

omeli [17]

Step-by-step explanation:

Given function is,

$y=55x+80$

Where,

y = amount paid to the gym in dollars

x = number of months

So when x = 0, y =55(0)+80 = 80

i.e when number of months is 0, you have to pay a fee of $80.

This is the one time fee to open a membership.

After that for each month you have to pay $55 for each months.

So 55 is multiplied with x.

3 0

2 years ago

A department store is having a sale in which every item is 38% off its original price. The regular price of a suit is $112. How

lara [203]

Answer: $42.56

Step-by-step explanation:

Percent discount in department store = 38%

The regular price of suit = $112.

Price taken off the regular price of the suit =

38% of $112

So, 38% of $112

= (38/100) x $112

= 0.38 x $112

= $42.56

(i.e the customer will pay $112 - $42.56 = $69.44 only for the suit)

Thus, the price taken off the regular price of the suit is $42.56

3 0

3 years ago

Verify identity: (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Nikitich [7]

So hmmm let's do the left-hand-side first

$\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

now, let's do the right-hand-side then

$\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

7 0

2 years ago