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olganol [36]
3 years ago
13

The size of the hermit crab’s shell depends on

Mathematics
2 answers:
elixir [45]3 years ago
5 0

Answer:

Length of crab in x years = x + 2

Step-by-step explanation:

Since the crab grows 1 inch per year, and it would grow:

1 inch in the 1st year

2 inches in the 2nd year

3 inches in the 3rd year

4 inches in the 4th year

.

.

.

And so on.

However, in order to find the total length, you need to remember that the crab is originally 2-inches long. So the total length after x years would be the number of inches it grew over the years (x) + the length before the growth (2).

Thus, Length of crab in x years = x + 2

steposvetlana [31]3 years ago
5 0
Length of crab in x years = x + 2
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(8y²)(-3x²y²)(2/3xy⁴)<br><br> HELP PLEASEEEEEEEE
IRINA_888 [86]

Step-by-step explanation:

1 Remove parentheses.

8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}

8y

2

×−3x

2

y

2

×

3

2

xy

4

2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

b

a

×

d

c

=

bd

ac

.

\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}

3

8y

2

×−3x

2

y

2

×2xy

4

3 Take out the constants.

\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(8×−3×2)y

2

y

2

y

4

x

2

x

4 Simplify 8\times -38×−3 to -24−24.

\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(−24×2)y

2

y

2

y

4

x

2

x

5 Simplify -24\times 2−24×2 to -48−48.

\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

−48y

2

y

2

y

4

x

2

x

6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

a

x

b

=x

a+b

.

\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}

3

−48y

2+2+4

x

2+1

7 Simplify 2+22+2 to 44.

\frac{-48{y}^{4+4}{x}^{2+1}}{3}

3

−48y

4+4

x

2+1

8 Simplify 4+44+4 to 88.

\frac{-48{y}^{8}{x}^{2+1}}{3}

3

−48y

8

x

2+1

9 Simplify 2+12+1 to 33.

\frac{-48{y}^{8}{x}^{3}}{3}

3

−48y

8

x

3

10 Move the negative sign to the left.

-\frac{48{y}^{8}{x}^{3}}{3}

−

3

48y

8

x

3

11 Simplify \frac{48{y}^{8}{x}^{3}}{3}

3

48y

8

x

3

to 16{y}^{8}{x}^{3}16y

8

x

3

.

-16{y}^{8}{x}^{3}

−16y

8

x

3

Done

6 0
2 years ago
Helppppppppppppppppppp please
Svetlanka [38]

Answer:

\sqrt[]{\frac{x+8}{4}}-3

Step-by-step explanation:

g(x)=4(x+3)^2-8

First rewrite g(x) as y

y=4(x+3)^2-8

Now swap y and x

x=4(y+3)^2-8

Add 8 on both sides.

x+8=4(y+3)^2-8+8

x+8=4(y+3)^2

Divide by 4.

\frac{x+8}{4} =\frac{4(y+3)^2}{4}

\frac{x+8}{4}=(y+3)^2

Extract the square root on both sides.

\sqrt[]{\frac{x+8}{4}}=\sqrt[]{(y+3)^2}

\sqrt[]{\frac{x+8}{4}}=y+3

Subtract 3 on both sides.

\sqrt[]{\frac{x+8}{4}}-3=y+3-3

\sqrt[]{\frac{x+8}{4}}-3=y

4 0
3 years ago
The square root of me and you is......<br> Brainliest for cutest answer!!
Alexandra [31]

Answer:

An adorable kitten!

Step-by-step explanation:

5 0
3 years ago
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Consider all angles whose reference angle is 45° with terminal sides not in Quadrant I.The angles that share the same cosine val
Oksana_A [137]

Solution:

As we know reference angle is smallest angle between terminal side and X axis.

As cosine 45 ° is always positive in first and fourth quadrant.

i.e CosФ, Cos (-Ф) or Cos(2π - Ф) have same value.

As, Cos 45°, Cos (-45°) or Cos ( 360° - 45°)= Cos 315°are same.

So, Angles that share the same Cosine value as Cos 45° have same terminal sides  will be in Quadrant IV having value Either Cos (-45°) or Cos (315°).

Also, Cos 45° = Sin 45° or Sin 135° i.e terminal side in first Quadrant or second Quadrant.



6 0
3 years ago
Read 2 more answers
READ the question and choose an answer below. You may look at the maps in Resources for Chapter 9, Lesson 1 and or you may use t
disa [49]
The answer is is Afghanistan
4 0
3 years ago
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