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2 years ago

Determine the value of x

Mathematics

2 answers:

DanielleElmas [232]2 years ago

6 0

Answer:Generally, the algebraic expression should be any one of the forms such as addition, subtraction, multiplication and division. To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Step-by-step explanation:

Dvinal [7]2 years ago

3 0

x = 2a

a = 3

x = 2 · 3 = 6 ← the end answer

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I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

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

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

Solve the equation plz and thx

kodGreya [7K]

It is A bro 4 eso a huevo

5 0

3 years ago

Read 2 more answers

Does the point (-4, 2) lie inside or outside or on the circle x^2 + y^2 = 25?

stiv31 [10]

Given equation of the Circle is ,

$\sf\implies x^2 + y^2 = 25$

And we need to tell that whether the point (-4,2) lies inside or outside the circle. On converting the equation into Standard form and determinimg the centre of the circle as ,

$\sf\implies (x-0)^2 +( y-0)^2 = 5 ^2$

Here we can say that ,

• Radius = 5 units

• Centre = (0,0)

Finding distance between the two points :-

$\sf\implies Distance = \sqrt{ (0+4)^2+(2-0)^2} \\\\\sf\implies Distance = \sqrt{ 16 + 4 } \\\\\sf\implies Distance =\sqrt{20}\\\\\sf\implies\red{ Distance = 4.47 }$

Here we can see that the distance of point from centre is less than the radius.

Hence the point lies within the circle .

5 0

3 years ago

Read 2 more answers

What is the missing number?

JulsSmile [24]

Answer:

y = 2^{x}

Step-by-step explanation:

Given the above data for x and y.

From the algebraic expression;

y = 2^{x}

We can deduce that the value of y is equal to two (2) raise to the power of x.

When x = 1, y = 2

y = 2^{x}

y = 2^{1}

y = 2

When x = 2, y = 4

y = 2^{x}

y = 2^{2}

y = 4

When x = 3, y = 8

y = 2^{x}

y = 2^{3}

y = 8

The above calculations can be used to determine the other values of y with respect to x.

3 0

3 years ago

How can you describe the graph of the equation ax + by = c? Explain

scoundrel [369]

Answer:

When A and B are not both zero, the graph of Ax + By = C is always a line. The graph of Ax + By = C, where A and B are not both zero, is a line. Divide both sides by B. Because the form Ax + By = C can describe any line, it is called the standard form of an equation for a line.

Hope this helps you!

6 0

2 years ago