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2 years ago

Solve for x 64^x^2 = 256^x

Mathematics

1 answer:

kifflom [539]2 years ago

8 0

Answer:

x = 0, 1.333333...

Step-by-step explanation:

1) Convert both sides to the same base.

$16^{1.5x^{2} } =16^{2x}$

2) Cancel the base of 16 on both sides.

$1.5x^{2} =2x$

3) Move all terms to one side.

$1.5x^{2} -2x=0$

4) Factor out the common term x.

$x(1.5x-2)=0$

5) Solve for x.

1 - Ask: When will x(1.5x-2) equal zero?

When x = 0 or 1.5x - 2 = 0
2 - Solve each of the 2 equations above.

x = 0, 1.333333...

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10(8x^2-8x)<br> 10(8x <br> 2<br> −8x)

Law Incorporation [45]

Answer: 6400<em>x²</em> · (<em>x - 1)²</em>

The rule says: To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x-1) and the exponents are:

1, as (x-1) is the same number as (x-1)1

And 1, as (x-1) is the same number as (x-1)1

The product is therefore, (x-1) (1 + 1) = (x-1)²

6 0

3 years ago

Read 2 more answers

Write an equation of the line in slope-intercept form

andre [41]

Answer:

y = 6

x = 3

Step-by-step explanation:

To find y and x on a graph, you must count how many squares one dot intercepts the same line as the other dot (in this case it's three) and for the y axis you do the ssme thing; find out how many squares are between both dots between the point where they intercept each other but for this time, it's vertically (in this case it's 6)

7 0

3 years ago

jekas [21]

option B is the answer of this question

3 0

2 years ago

Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from

Lapatulllka [165]

<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$ </h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

$x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8$

The area of a triangle is= $\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]$

= $|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|$

= $|-54|$ = 54 square units

The length of AC = $\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}$

= $\sqrt{(2-12)^{2} +(1-8)^2}$

= $\sqrt{149}$ units

Let the perpendicular distance from B to AC be = x

According To Problem

$\frac{1}{2} \times x \times \sqrt{149} = 54$

⇔ $x =\frac{108}{\sqrt{149} }$ units

Therefore the perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$

5 0

3 years ago

Four years after opening an account that paid simple interest of 4.25% annually, a depositor withdrew the $2,380 in interest ear

satela [25.4K]

Answer:

Step-by-step explanation:

In order to figure out how much money was left in the account after the interest was withdrawn, we have to first find out how much money was initially deposited to earn that amount of interest! The means to find that initial investment is found in the simple interest formula

prt = I, where

p is the initial investement,

r is the interest rate in decimal form,

t is the time in years, and

I is the interest earned. Notice that we have all those things but the p.

Filling in:

p(.0425)(4) = 2380 and

.17p = 2380 so

p = 14000

That means that 14000 was initially invested. If the depositor withdrew the 2380, then

14000 - 2380 is the amount left in the account, namely, $11620

7 0

3 years ago