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3 years ago

Use the distriputive property to rewrite the expression. 11 (4d+6)

Mathematics

2 answers:

amm18123 years ago

8 0

Essentially 11(4d + 6) = 11*4d + 6*11

44d + 66

DanielleElmas [232]3 years ago

8 0

11(4d + 6)

Distribute (multiply) 11 inside the parentheses.

(11 * 4d) + (11 * 6)

44d + 66 is your answer.

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If the diagonal of a square is approximately 12.73, what is the length of its sides?

Alja [10]

Answer:

The length of the sides of the square is 9.0015

Step-by-step explanation:

Given

The diagonal of a square = 12.73

Required

The length of its side

Let the length and the diagonal of the square be represented by L and D, respectively.

So that

D = 12.73

The relationship between the diagonal and the length of a square is given by the Pythagoras theorem as follows:

$D^{2} = L^{2} + L^{2}$

Solving further, we have

$D^{2} = 2L^{2}$

Divide both sides by 2

$\frac{D^{2}}{2} = \frac{2L^{2}}{2}$

$\frac{D^{2}}{2} = L^2$

Take Square root of both sides

$\sqrt{\frac{D^{2}}{2}} = \sqrt{L^2}$

$\sqrt{\frac{D^{2}}{2}} = L$

Reorder

$L = \sqrt{\frac{D^{2}}{2}}$

Now, the value of L can be calculated by substituting 12.73 for D

$L = \sqrt{\frac{12.73^{2}}{2}}$

$L = \sqrt{\frac{162.0529}{2}}$

$L = \sqrt{{81.02645}$

$L = 9.001469325$

$L = 9.0015$ (Approximated)

Hence, the length of the sides of the square is approximately 9.0015

7 0

3 years ago

Find the length of a square with an area of 169 in2.

mrs_skeptik [129]

Find the length of a side of a square with an area of 169 in^2.

Answer:

D. 13 in

Step-by-step explanation:

A square has sides of equal length.

A = L^2 where: A = area and L = side

L^2 = 169

L=√169

L=13 in^2.

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3 years ago

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Ilia_Sergeevich [38]

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4 years ago

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irga5000 [103]

Answer:

$4c^{7}d^{13}$

Step-by-step explanation:

$(2cd^{4} )^{2} *(cd)^{5}$

$4c^{2}d^{8} *c^{5} d^{5}$

$4c^{7}d^{13}$

4 0

3 years ago

Use induction to prove that 2? ?? for any integer n>0 . Indicate type of induction used.

Hoochie [10]

Answer with explanation:

The given statement is which we have to prove by the principal of Mathematical Induction

$2^{n}>n$

1.→For, n=1

L H S =2

R H S=1

2>1

L H S> R H S

So,the Statement is true for , n=1.

2.⇒Let the statement is true for, n=k.

$2^{k}>k$

---------------------------------------(1)

3⇒Now, we will prove that the mathematical statement is true for, n=k+1.

$\rightarrow 2^{k+1}>k+1\\\\L H S=\rightarrow 2^{k+1}=2^{k}\times 2\\\\\text{Using 1}\\\\2^{k}>k\\\\\text{Multiplying both sides by 2}\\\\2^{k+1}>2k\\\\As, 2 k=k+k,\text{Which will be always greater than }k+1.\\\\\rightarrow 2 k>k+1\\\\\rightarrow2^{k+1}>k+1$

Hence it is true for, n=k+1.

So,we have proved the statement with the help of mathematical Induction, which is

$2^{k}>k$

3 0

3 years ago