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Verizon [17]
2 years ago
5

Type the expressions as radicals. Y^5/2

Mathematics
1 answer:
Lapatulllka [165]2 years ago
7 0

Answer:

it will be square root y^5

Step-by-step explanation: Apply the rule x m/n=n√x m to rewrite the exponentiation as a radical.

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gavmur [86]
Well, what are the pictures??
5 0
3 years ago
Help me now please please
Nikolay [14]

Answer:

Step-by-step explanation:

What you would want to do is add the $15.75 you would want to add it 5 times

8 0
3 years ago
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The correct and best answer will be marked as brainiest
marta [7]

Answer:

x=10

m=3

Step-by-step explanation:

The angles are the same since the sides are the same length (isosceles triangle)

55 = 5x+5

Subtract 5

55-5 =5x+5-5

50 = 5x

Divide by 5

50/5 = 5x/5

10=x

The altitude is a perpendicular bisector so

5m-3 = 2m+6

Subtract 2m from each side

5m-3-2m = 2m+6-2m

3m-3 = 6

Add 3 to each side

3m-3 +3 =6+3

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Divide by 3

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8 0
3 years ago
I need this question solved with derivatives please:
Shkiper50 [21]

Answer:

16,242. 7 cm^3.

Step-by-step explanation:

We need to cut off a square piece at the 4 corners of the cardboard.

Let the length of their edges be x cm.

The volume of the box will be:

V = height * width * length

V = x(100-2x)(40-2x)

V = x(4000 - 200x - 80x + 4x^2)

V = x(4x^2 - 280x + 4000)

V =  4x^3 + - 280x^2 + 4000x

Finding the derivative:

dV / dx =  12x^2 - 560x + 4000  = 0  ( when  V is a maxm or minm.)

4(3x^2 - 140x + 1000) = 0

x =  37.86, 8.80.

Looks like x = 8.80 is the right value but we can check this out be looking at the sign of the second derivative:

V" =  24x - 560, when x = 8.8   V" is negative so this is a Maximum for V.

So the maximum volume of the box is when x = 8.8 so we have

V =  8.8(100-2(8.8)(40 - 2(8.8)

=  16,242. 7 cm^3.

5 0
3 years ago
For this figure, what is the value of x?
andrew11 [14]
The angles inside a triangle always add up to 180 degrees.
The whole thing is one big triangle made of two smaller ones.

28 + 58 + 77 + x = 180
163 + x = 180
x = 180 - 163
x = 17
5 0
3 years ago
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