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Scrat [10]
3 years ago
15

Which monomial is a perfect cube? 16x6 27x8 32x12 64x6

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0
16x16 is the correct Answer
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Solve the system by substitution.<br> - 2x – 10y = -2<br> -4y = x
mihalych1998 [28]

Answer:

x = -4 and y = 1

Step-by-step explanation:

hope this helps please like and mark as brainliest

4 0
2 years ago
Which of the following functions is quadratic?
lions [1.4K]

if u look at it and and which ones have more then u will get ur answer

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2 years ago
Solve the following system of linear equations algebraically.<br> 3x + y = 5<br> 5x – 4y = -3
Korvikt [17]

3x+y=5 & 5x-4y=-3

3x+5x=8x+y-4y=-3

           8x+y-4y=-3

     8x+3y=-3

         -3     -3

8x=-6

3 0
2 years ago
Read 2 more answers
Math problem+Zoey, a Gelada Baboon, is tied to one of the outside corners of a regular hexagon-shaped research facility, where h
dexar [7]

Answer

The monkey will be able to cover an area of 22.86 meters which is equivalent to two sides and half of the building.

Step-by-step explanation:

The length of the monkeys rope = 75 feet long

Each side of the hexagon shaped facility = 30 feet

Note that, 1 feet = 0.3048 m

if we convert 30 feet to Meter, we have,

30 × 0.3048 = 9.144

Therefore, 30 + 30 +15 = 75 which is equivalent to the two side and half side of the hexagon shaped building

75 feet × 0.304 meter = 22.86 meters.

So the total area the monkey will be able to cover will be

= 22.86 Meters which is equivalent to two sides and half of the hexagon shape building.

8 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
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