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3 years ago

Which inequality is represented by this graph?

Mathematics

1 answer:

avanturin [10]3 years ago

4 0

Answer:

The Option D (x ≥ - 53) is correct for the given graph.

Step-by-step explanation:

As shown in the graph blue part is from - 53 to -50 including -53.

therefore x ≥ - 53.

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Help Por favor! Will mark brainliest!

postnew [5]

Answer: An adult’s ticket costs $9 while a children’s ticket costs $5.

Step-by-step explanation:

1). 2x + 3y = 33

2). 5x + 2y = 55

3). I’m going to use substitution.

Isolate a variable (x):

2x + 3y = 33

2x = -3y + 33

X = (-3y + 33)/2

X = -3/2y + 33/2

Substitute and solve for y:

5x + 2y = 55

5(-3/2y + 33/2) + 2y = 55

-15/2y + 165/2 + 2y = 55

-11/2y = -55/2

Y = 5 <— children’s ticket.

Solve for x:

2x + 3y = 33

2x + 3(5) = 33

2x + 15 = 33

2x = 18

X = 9 <— adult’s ticket.

Check:

5x + 2y = 55

5(9) + 2(5) = 55

45 + 10 = 55

55 = 55

2x + 3y = 33

2(9) + 3(5) = 33

18 + 15 = 33

33 = 33

8 0

3 years ago

What equality is shown here on the number above?

grigory [225]

Answer:

the equality shown is : x is greater than or equal to -3.

6 0

3 years ago

What is 4 times 2 please I’m in need of an answer

MAXImum [283]

8 is the answer to your question

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2 years ago

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HELP PLZ ASPA 100 POINTS WILL GIVE BRAINLIEST!!!

Varvara68 [4.7K]

Step-by-step explanation:

(x10,y-2), reflection over y=1

6 0

2 years ago

Read 2 more answers

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

3 years ago