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2 years ago

Last question I swear

Mathematics

1 answer:

kirza4 [7]2 years ago

3 0

Answer:

y = 3

Step-by-step explanation:

Subtract 11 from both sides to get:

2y = 6

Then divide both sides by 2 to isolate y:

y = 3

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What is 5.37 as a mixed number.

Gnom [1K]

Answer: 5.37 as a mixed number is 5 37/100

To write a decimal number as a mixed number, you must separate the whole part of the decimal part:
5.37 = 5 + 0.37 = 5 0.37
And express the decimal part as a fraction:
0.37=37/100
Then:
Answer: 5.37 as a mixed number is 5 37/100

7 0

2 years ago

In what year will the two quantities be equal?

Galina-37 [17]

Its the second because i did this already

3 0

3 years ago

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.

case a) $x^{2}+2x+4=0$

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

case b) $x^{2}-2x+4=0$

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

case c) $x^{2}+2x-4=0$

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

case d) $x^{2}-2x-4=0$

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

the answer is

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

5 0

2 years ago

Read 2 more answers

What is an example of when you would want consistent data and, therefore, a small standard deviation?

steposvetlana [31]

Answer:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

Step-by-step explanation:

For example if we have the following data:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

We see that the data are similar for all the observations so we would expect a small standard deviation

If we calculate the sample mean we can use the following formula:

$\bar X=\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

8 0

2 years ago

The product of two positive integers is 120. One number is 2 less than five times the other. Select the equation that can be use

svetlana [45]

lets assume one number is x

then 2nd number will be 5x-2

so product is 120

x (5x-2) = 120

5x^2 - 2x = 120

so answer is last option

8 0

3 years ago

Read 2 more answers