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3 years ago

What is 54386 + 5697 togther?

Mathematics

2 answers:

Elina [12.6K]3 years ago

7 0

<span>54386 + 5697 togther is </span>60083

snow_lady [41]3 years ago

5 0

The answer is 60,083.

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What number is 4 time another. The sum of the reciprocals is 15/4. find the numbers

antoniya [11.8K]

Answer:

3/4 , 3

Step-by-step explanation:

Let one number = x

Other = 4x

Sum:

x + 4x = 15/4

5x = 15/4

x = 3/4

Thus the numbers are 3/4 , 3

8 0

3 years ago

Put the integers in least to greatest -6, 3, -12, -8, 15, -13, 4, 10, -14, 2, 11

Strike441 [17]

Answer:

Step-by-step explanation:

-14,-13,-12,-8,-6,2,3,4,10,11,15

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4 years ago

Read 2 more answers

Cecilia's train travled at 240 miles in 3 hours. How far will it travel in 5.5 hours.

yanalaym [24]

5.5<span>÷3 is 1.83333333, that times 240 is 440</span>

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4 years ago

Uhhh I am struggling with this question algebra 1

Olin [163]

Answer:

-2x^2 +16x -8

Step-by-step explanation:

Oh s-

Like terms i guess Same variable you can substract. Remember you only substract from 1 number not from all.

-2x^2 stays the same as there is no like term

9x - (-7x)

-2 - (6)

-2x^2 + 9x+ 7x -2-6

-2x^2 +16x -8

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3 years ago

Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

6 0

4 years ago