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2 years ago

12

Help me plzzzzzzzzzzzzzzzzzzz

1 answer:

elena55 [62]2 years ago

4 0

Answer:

-5

Step-by-step explanation:

1*1=1

1-6= -5

Hope this helped!

You might be interested in

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

2 years ago

HELP ASAP PLEASE!!!

xenn [34]

Answer:

E. None of these

Step-by-step explanation:

Points: (11, -4), (13, -7)

slope = m = (y2 - y1)/(x2 - x1) = (-7 - (-4))/(13 - 11) = (-7 + 4)/2 = -3/2

y - y1 = m(x - x1)

y - (-4) = (-3/2)(x - 11)

y + 4 = (-3/2)x + 33/2

y + 8/2 = (-3/2)x + 33/2

y = (-3/2)x + 25/2

Answer: E. None of these

4 0

2 years ago

Read 2 more answers

Help with this one please

den301095 [7]

point R ((:

reason being (-5,-6) over the y axis is (5, -6)

3 0

3 years ago

Pick two that are NOT equivalent to 6(5x + 4) - 10

bonufazy [111]

Answer:

a and c

Step-by-step explanation:

it would be a and c because the first expreesion in the sentace equals 54

<u><em>A and C are the only options that do not equal ( They both equal 24 not 54!) </em></u>

3 0

2 years ago

Read 2 more answers

What is the least positive integer divisible by the numbers 2, 4, and 7?

Viefleur [7K]

Answer:

Least positive integer divisible by the numbers 2, 4, and 7 is 28

Step-by-step explanation:

We can find the least positive integer divisible by the numbers 2, 4, and 7 by taking the LCM

First lets List all prime factors for each number.

Prime Factorization of 2

2 is prime => $2^1$

Prime Factorization of 4 is:

2 x 2 => $2^2$

Prime Factorization of 7 is:

7 is prime => $7^1$

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 7

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 7 = 28

7 0

3 years ago