All categories

3 years ago

Two less than the product of 8 and a number is 8

Mathematics

1 answer:

Eva8 [605]3 years ago

3 0

Answer:

x=1.25

Step-by-step explanation:

Let us replace the number with the variable "x". You will then get this equation:

8x-2=8

Move the -2 to 8 and you would get the equation:

8x=8+2

Simplify:

8x=10

Simplify:

x=10/8

You would then get the answer "x=1.25"

I hope this helps :)

You might be interested in

How are u supposed to find the area only with the circumference?

seraphim [82]

Answer:

We can find the radius from circumference and then area

Circumference(C) = 2πr

r = C/2π

r = 20/2π

r = 10/π

Area = πr²

Area = π × 10/π × 10/π

Area = 100/π

Area = 100/3.14

Area = 10000/314 = 31.84 cm² (approx)

Area = 32 cm² (Rounding to nearest whole number)

6 0

3 years ago

Read 2 more answers

The question is in the picture, please help.

beks73 [17]

Answer:

b c and f edit: also A

Step-by-step explanation:

just gotta find what is reflexive

3 0

3 years ago

What is the answer for 0.9 divided by 10 to the power of 3

velikii [3]

Answer:

0.0009

Step-by-step explanation:

0.9 / 10^3 = 0.0009

5 0

3 years ago

Is this right if not please tell me the right answer

zhuklara [117]

Answer: 57% for the first and 41% for the second

Step-by-step explanation:

Treatment 1 consists of 40 people and 23 had improved health take 23 and divide it by 40 and you get 0.575 multiply that by 100 and you get 57% do the same for the second question and you get 0.416 multiply by 100 and you get 41%

8 0

3 years ago

SHOW YOUR WORK, <br><br>PLEASE HELP!!!!!!!!!

Shkiper50 [21]

To find the inverse, we swap the variables y and x, then solve for the new y.

3a. $y=\frac{3}{x-1}$

Swapping the variables: $x=\frac{3}{y-1}$
Solving for y: $x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}$
The domain of this inverse is $x ≠ 0$ .
3b. $y=x^2-1$

Swapping: $x = y^2 - 1$
Solving for y: $y^2 = x + 1 \\ y = \sqrt{x+1}$
The domain of this inverse is $x ≥ -1$ .
3c. $y=\sqrt[3]{\frac{x-7}{3}}$
Swapping: $x=\sqrt[3]{\frac{y-7}{3}}$
Solving for y: $x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7$
The domain of this inverse is all real numbers.
4a. $y=\frac{3}{x-1}$ , $y=1+\frac{3}{x}$
$y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x$
$y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x$

4c. $y=\sqrt[3]{\frac{x-7}{3}}$ , $y=3x^3+7$
$y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x$
$y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x$

3 0

3 years ago