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bekas [8.4K]
3 years ago
5

A retirement community has a sign at its main entrance that says "Residents over the age of 55 welcome." Which of the following

inequalities best represents the age of residents welcome in the retirement community?
z > 55
z < 55
z ≥ 55
z ≤ 55
Mathematics
2 answers:
IrinaVladis [17]3 years ago
4 0
My best answer is z<55 hope this helps
horsena [70]3 years ago
4 0

Answer:

the answer is the first one z>55

Step-by-step explanation:

I did this exam on got the question right

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If you invest $823 at an interest rate of 3 percent,how much money will you have in total afer nine years?​
valentinak56 [21]

Answer:

$1,045.21

Step-by-step explanation:

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3 years ago
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A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six
kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

V = Length \times Width \times Height \\\\

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

Width = 3 - 2x \\\\

The length of the shaded region is given by

Length = \frac{1}{2} (5 - 3x) \\\\

So, the volume of the box becomes,

V =  \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V =  \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V =  \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V =  \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

a = 18 \\\\b = -38 \\\\c = 15 \\\\

x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 +  19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\

Volume of the box at x= 1.59:

V =  \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\

Volume of the box at x= 0.53:

V =  \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0
3 years ago
What is the measure of x?
Sveta_85 [38]

Answer:

Hello!

After reviewing the problem you have provided I have come up with the correct solution:

x= 9

Step-by-step explanation:

To come up with this solution you have to first realize that the smaller triangle is a proportionally scaled down version of the entire larger triangle! (I will show what I mean in a linked picture)

So after we have realized that the smaller triangle is a scaled down version of the larger one, we can then create a formula or ratio to calculate the value of the missing side of the larger triangle (being x+6=??).

To create the formula/ratio I divided 10inches by 4inches. Thus the larger triangle is 2.5 times larger than the smaller one.

I then use this ratio to figure out the missing length of the larger triangle by doing:

6inches x 2.5 = 15inches.

I then inputed the 15inches into the formula of the missing side:

x+6=15

Subtracted 6 from both sides to simplify, and came up with the solution!

x=9

Let me know if this helps!

4 0
2 years ago
If B=3x^{2}-x+3 and A=x-6, find an expression that equals 2B+3A in standard form.
Ivan

9514 1404 393

Answer:

  6x^2 +x -12

Step-by-step explanation:

Substitute for A and B and collect terms.

  2B +3A

  = 2(3x^2 -x +3) +3(x -6) . . . . substitute for A and B

  = 6x^2 -2x +6 +3x -18 . . . . . eliminate parentheses

  = 6x^2 +x -12 . . . . . . . . . . . . collect terms

7 0
3 years ago
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