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4 years ago

How would you find out what 1/11 is as a decimal using a calculator?

Mathematics

2 answers:

Elanso [62]4 years ago

5 0

Answer:

a the first one

Step-by-step explanation:

1/11 means 1 divided by 11

Lisa [10]4 years ago

4 0

Answer:

A. Punch in "1 divided by 11"

Step-by-step explanation:

You might be interested in

A rectangle that is 2 inches by 3 inches has been scaled by a factor of 7 if you want to cooy back to its original size by what

slamgirl [31]

Answer:

1/7

Step-by-step explanation:

Given that:

Using a scale factor of 7 ; the dimension of a rectangle is : 2 inches by 3 inches

In other to obtain the original size of the rectangle, all that is needed is to multiply the sacl d dimension by the reciprocal of the initi. Scale factor used, that is 1 / 7

This gives :

(2 inches by 3 inches) * 1/7

2/7 inches by 3/7 inches

Hence, required scale factor is 1/ 7

3 0

3 years ago

Solve the following system. 1 v=()x2 + 2* 2 + 2x-1 and 3x-y=1

icang [17]

Answer:

x = 1 + i, x = 1 - i

Step-by-step explanation:

Make the second equation y = 3x - 1

Substitute: 3x - 1 = 1/2x^2 + 2x

Simplify: 6x - 2 = x^2 + 4x

x^2 + 4x + 2 = 6x

x^2 - 2x + 2 = 0

Complete the square: x^2 - 2x + 1 = - 1

(x - 1)^2 = -1

x - 1 = ± √-1

Take out the negative: x - 1 = ± i

x = 1 + i, x = 1 - i

i is for imaginary number

8 0

3 years ago

Read 2 more answers

israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

4 years ago

6. Evaluate the following expression: 10 + 4(6-2) А. 56 B. 82 С. 18 D. 26

irakobra [83]

Answer:

D. 26

Step-by-step explanation:

$10+4(6-2)\\10+4(4)\\10+16\\26$