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2 years ago

Slove it 3x-5=x+3 :)))))))

Mathematics

2 answers:

juin [17]2 years ago

6 0

Answer:

x=4

Step-by-step explanation:

3x-5=x+3

-x +5 -x +5

2x=8

next you divide 2 from 8

x=4

lidiya [134]2 years ago

5 0

Answer:

x=4

Slove it 3x-5=x+3

3x-5=x+3

3x-x-5=3

2x-5=3

2x=3+5

2x=8

x=4

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1 year ago

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The circle described by the equation <img src="https://tex.z-dn.net/?f=%28x-3%29%5E%7B2%7D%20%2B%28y%2B4%29%5E%7B2%7D%3D9" id="T

ElenaW [278]

Answer:

The point at which center of the image circle located is (-2, -3) so second option is correct.

Step-by-step explanation:

Given equation of circle is:

$(x-3)^{2} + (y+4)^{2} = 9$ (i)

The general equation of circle is mentioned below,

$(x-h)^{2} + (y-k)^{2} = r^{2}$ (ii)

Here 'r' represents the radius of the circle and (h,k) shown the center of the circle.

By comparing equation (i) and equation (ii), we get

r^2 = 9

r = 3

$(x-3)^{2} = (x-h)^{2}$

$(x-3) = (x-h)$

$h = 3$

$(y+4)^{2} = (y-k)^{2}$

$(y+4) = (y-k)$

$k = -4$

So the center of given circle is (h,k) = (3,-4)

Also, the circle is translated 5 units left, that is towards the -x-axis. Therefore h = 3 - 5 = -2

Also, the circle is translated 1 unit up, that is towards the +y-axis. Therefore k = -4 + 1 = -3

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2 years ago

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

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3 years ago