Ask question

Ask question

All categories

3 years ago

13

X-3 divided by 4 = 12

2 answers:

andreev551 [17]3 years ago

7 0

Answer:

$\Huge \boxed{x=51}$

$\rule[225]{225}{2}$

Step-by-step explanation:

$\displaystyle \frac{x-3}{4} =12$

Multiplying both sides by 4.

$\displaystyle \frac{x-3}{4} *4=12 *4$

$x-3=48$

Adding 3 to both sides.

$x-3+3=48+3$

$x=51$

$\rule[225]{225}{2}$

yulyashka [42]3 years ago

4 0

Answer:

x = 51

Step-by-step explanation:

(x-3) /4 = 12

Multiply each side by 4

(x-3) /4 *4 = 12*4

x-3 = 48

Add 3 to each side

x-3+3 = 48+3

x = 51

You might be interested in

Polynomial in standard form.

nataly862011 [7]

Answer:

-2x² - 8x - 15

Step-by-step explanation:

-(2x² + 4x + 11) + (-4)

-2x² - 8x - 11 - 4

-2x² - 8x - 15

6 0

3 years ago

Help plz!!!!!?!!!?!??!????

aleksandrvk [35]

700ml=23.6698fl
64-23.6698=40.3302
40.3302fl

5 0

2 years ago

A home mortgage is usually borrowed for how long?

lawyer [7]

<span>A home mortgage is usually borrowed for how long? The answer is D 20-30 years. A traditional home mortgage is borrowed for 30 years. In some cases this can be shorter or longer but the typical mortgage is between 20-30 years.</span>

4 0

3 years ago

Read 2 more answers

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

Gerald has a punch bowl in the shape of a hemisphere as shown below.

zmey [24]

The closest to the maximum number of cups the punch bowl can hold is 30

<h3>How to determine the number of cups?</h3>

The given parameters are:

1 cup = 15 cubic inches

Diameter of bowl, d = 12 inches

The radius is the half of the diameter.

So, we have:

r = 6

The volume of the bowl is then calculated using:

$V = \frac{2}{3}\pi r^3$

This gives

$V = \frac{2}{3}\pi * 6^3$

Evaluate

V = 452.16

The maximum number of cups is then calculated using:

Cups = 452.16/15

Evaluate

Cups = 30.1444

Approximate

Cups = 30

Hence, the closest to the maximum number of cups the punch bowl can hold is 30

Read more about volumes at:

brainly.com/question/1972490

#SPJ1

6 0

2 years ago