All categories

3 years ago

A parallelogram has an interior angle measuring 81º. Which MUST also be the measure of an interior angle of the

Mathematics

1 answer:

ANEK [815]3 years ago

8 0

Answer:

D)99 degrees

Step-by-step explanation:

The answer is D. All the angles in any quadrilateral add up to 360 degrees. And since the figure is a paralledlogram, for each angle there is another angle with the same measurement. So 81 + 81 + x + x = 360 which gives us 162 + 2x = 360. If you solve for x you get 99 degrees.

You might be interested in

there are 250 toothpicks in a regular size box. a jumbo size box is made by tripling the length,width, and height of the regular

MrMuchimi

Answer:6750

Step-by-step explanation:

8 0

4 years ago

Who many symmetry lines does this figure have

Nana76 [90]

Answer:

4 maybe

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help me 1 2 3 or 4

zepelin [54]

Answer:

You get 3 for solving this inequality.

(ANSWER IS NUMBER 2)

Step-by-step explanation:

Solve this like a normal equation.

2x is less that 9-3

2x is less than 6

x is less than 3

7 0

3 years ago

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

Zarrin [17]

Answer:

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

Therefore, x = 0.53 cm and the Maximum volume = 1.75 cm³

Step-by-step explanation:

Please refer to the attached diagram:

The volume of the box is given by

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

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

The width of shaded region is given by

$Width = 3 - 2x \\\\$

The length of shaded region is given by

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

So, the volume of the box becomes,

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

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 may solve the quadratic equation using the quadratic formula.

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

The values of coefficients a, b, c are

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

Substituting the values into quadratic formula yields,

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