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3 years ago

I’ve been stuck on this for a few minutes. I’m not really sure what to do..

Mathematics

1 answer:

rewona [7]3 years ago

7 0

Answer:

Step-by-step explanation:

You know that the three angles of a triangle add up to 180°. If one angle is 38° and another angle is 67°, then the third angle must be 180°-38°-67° = 75°.

You have a straight line across the bottom, and three angles meeting in the middle of it. Those three angles add up to 180°. One of them is 75°, another is 50°, so the third angle must be 180°-75°-50° = 55°.

That leaves you with one unknown angle. ? = 180°-40°-55° = 85°

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Looking at the triangle we see the angles:

58, x

We also see a right angle, therefore you now have:

90, 58, 0

All three angles of a triangle should have a total of 180 degrees, so we add the known angles and subtract the total from 180.

90 + 58 = 148

180 - 148 = 32

Therefore, x = 32 or C

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2 years ago

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3 years ago

If you add the year of my birth to this year, subtract the year of my 20th birthday and the year of my 85th birthday, and then a

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3 years ago

-8x - 11x<br> I really need help

Tema [17]

Answer: -19x

Step-by-step explanation: In this problem, we're asked to simplify the expression.

The two parts of this expression, -8x and 11x are called terms. The numbers in front of the variables are called coefficients.

Because the variables, x and x are identical, the terms get a special name and they're called like terms.

To subtract like terms, we simply subtract their coefficients and add the variable on to the difference of the two numbers.

Since -8 (-11) is -19, -8x - 11x simplifies to -19x.

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2 years ago

An open box with a square base is constructed so that it has a volume of 64in^3. The base needs to be made out of a material tha

Ivahew [28]

The volume of a box is the amount of space in the box

The dimensions that minimize the cost of the box is 4 in by 4 in by 4 in

<h3>How to determine the dimensions that minimize the cost</h3>

The dimensions of the box are:

Width = x

Depth = y

So, the volume (V) is:

$V = x^2y$

The volume is given as 64 cubic inches.

So, we have:

$x^2y = 64$

Make y the subject

$y = \frac{64}{x^2}$

The surface area of the box is calculated as:

$A =x^2 + 4xy$

The cost is:

$C = 2x^2 + 4xy$ --- the base is twice as expensive as the sides

Substitute $y = \frac{64}{x^2}$

$C = 2x^2 + 4x * \frac{64}{x^2}$

$C =2x^2 + \frac{256}{x}$

Differentiate

$C' =4x - \frac{256}{x^2}$

Set to 0

$4x - \frac{256}{x^2} = 0$

Multiply through by x^2

$4x^3 - 256= 0$

Divide through by 4

$x^3 - 64= 0$

Add 64 to both sides

$x^3 = 64$

Take the cube roots of both sides

$x = 4$

Recall that:

$y = \frac{64}{x^2}$

So, we have:

$y = \frac{64}{4^2}$

$y = 4$

Hence, the dimensions that minimize the cost of the box is 4 in by 4 in by 4 in