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4 years ago

7

What set does 13.444 belong to

1 answer:

sergij07 [2.7K]4 years ago

5 0

Answer:

13.4445 I think.

Step-by-step explanation:

You might be interested in

A retired math teacher takes her niece shopping. She tells her niece that they can spend x dollars and

pashok25 [27]

Answer:

Division. The niece can divide both sides of the equation by 19 to find the answer.

Let me know if this helps!

8 0

3 years ago

Two-thirds a number plus 4 is 7

ICE Princess25 [194]

The equation (form) is: $\frac{2}{3}x + 4 = 7$

The solution of the equation is: $x = \frac{9}{2}$

Explanation:

First, let's call the number we are going to solve for "x".

Then two thirds of this number can be written as:

$\frac{2}{3}x$

Then "plus 4" means add 4 we get:

$\frac{2}{3}x + 4$

"is 7" means that the above expression is equal to 7 (as follows):

$\frac{2}{3}x + 4 = 7$

The solution of the above equation is:

$\frac{2}{3}x + 4 = 7 \\ \frac{2}{3}x = 3 \\ x = \frac{9}{2}$

6 0

3 years ago

Katen [24]

The error is in the second step they subtracted 3 from both sides but it was a 3x, which you can not do.
Correction.
9x+18+3x=1
12x+18=1
12x=-17
x=-17/12

3 0

3 years ago

Read 2 more answers

What is 3x4 when it is divided by 4

Anna71 [15]

Answer:

3

Step-by-step explanation:

3*4/4

4/4 is equal to 1

3*1

3

5 0

3 years ago

Read 2 more answers

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

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

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

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

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers