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3 years ago

????????? Answer ?????

Mathematics

2 answers:

N76 [4]3 years ago

8 0

I think that the answer is A.

kiruha [24]3 years ago

7 0

I'm pretty sure the answer is A.

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What does it mean to be "certain" to occur?

Luden [163]

Certain means the event will happen.

Impossible means the event would never happen.

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3 years ago

Simplify the following

kondaur [170]

The answer is z^2+3z+1

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3 years ago

What are a, b, c and d

GuDViN [60]

Answer:

Step-by-step explanation:

4 g.

c=42

a=b=d

a+b+42=180

a+a=180-42

2a=138

a=138/2=69

a=69

b=69

c=42

d=69

5a.

a=30

angle O=180-(a+30)=180-(30+30)=180-60=120°

d=80

angle O=180-(d+80))=180-(80+80)=180-160=20°

central angle O of middle triangle=180-(120+20)=180-140=40°

b=c

b+c+40=180

b+b=180-40

2b=140

b=140/2=70

b=70°

c=70°

7 0

2 years ago

PLS HELP I WILL MARK BRAINLIEST 6TH GRADE LEVEL

Liula [17]

Answer:

1. A 2. B

Step-by-step explanation:

1. 9(6 times x) and 9 x 6 + 9 times x

Thhis is correct because you do the same for both

2. 1 pound is 16 ounces. So 12.5 x 16 = 200

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3 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