All categories

2 years ago

Someone know how to do this Im very confusing about how to do it

Mathematics

1 answer:

Sidana [21]2 years ago

3 0

It would be e, because it went down 10 instead of 5

You might be interested in

How do I solve out 5x188

mihalych1998 [28]

Answer:

940

Step-by-step explanation:

188

940

--------------------------------------------------------

8 * 5 = 40

Carry the 4 to the next 8

8 * 5 = 40

Add the 4 to the 40

40 + 4 = 48

Carry the 4 to the 1

1 * 5 = 5

Add the 4

5 + 9

= 940

*Hope this helps .^.*

5 0

3 years ago

Consider the square root function f(x) = sqrt x + 2 + 1 to complete the table of values below. find each missing value and then

Scrat [10]

Given the function :

$f(x)=\sqrt[]{x+2}+1$

We need to find each missing value

Given x = -3 , -2 , -1 , 2 , 7

So, substitute with each value of x to find the corresponding value of f(x)

$x=-3\rightarrow f(x)=\sqrt[]{-3+2}+1=\sqrt[]{-1}+1$

So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)

$\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\ \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\ \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\ \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}$

the graph of the function and the points will be as shown in the following image :

5 0

1 year ago

I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago

Complete the square: -3x^2 + 30x -52 express in vertex form

nadezda [96]

the answer to this is

the two is for the top of the five 2

-3(x - 5) +23

4 0

3 years ago

U-3/4=4 please help me ty

Ne4ueva [31]

To solve the equation:

$\frac{u - 3}{4} = 4 \\ \frac{u - 3}{4} \times 4 = 4 \times 4\\ u - 3 = 4 \times 4 \\ u - 3 + 3 = 16 + 3 \\ u = 16 + 3 \\ u = 19$

Therefore, the answer is u=19.

Hope it helps!

8 0

3 years ago

Read 2 more answers