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3 years ago

Write the expression in the simplest form. (-11/2 x +3) -2(- 11/4 x - 5/2)

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

-11x+8

Step-by-step explanation:

I did this quickly but you would just multiply the right side by -2 and add like terms.

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Question 1 GIVING BRAINLIESTT!!!!!!!!

Rama09 [41]

Answer is (-1.5,5)
Midpoint formula is M= ((X1+X2)/2), ((Y1+Y2)/2)
So it becomes ((-4+1)/2), ((9+1)/2)= (-1.5,5)

3 0

2 years ago

Help me find the answer

emmainna [20.7K]

Answer:

2.30 = ?

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp side/ hypotenuse

sin 50 = ? / 3

Multiply each side by 3

3 sin 50 = ?

2.298133329 = ?

To the nearest hundredth

2.30 = ?

5 0

2 years ago

Find the inverse of the relation (-3,4), (-1,0), (6,0).

allochka39001 [22]

To find the inverse of a relation, we switch the x and y values in each point.

So the inverse would be {(4, -3), (0, -1), (0, 6).

7 0

3 years ago

Read 2 more answers

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

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

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