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3 years ago

What is the coefficient in the following expression?

Mathematics

2 answers:

tensa zangetsu [6.8K]3 years ago

8 0

C ) 5! hope that helped

ikadub [295]3 years ago

4 0

Answer:

c)5

Step-by-step explanation:

Coefficient is always the numerical value before the variable

I hope this helped and have a good rest of your day!

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The radius of a cylindrical construction pipe is 1.5ft. If the pipe is 18ft long, what is its volume? Use the value 3.14 for π ,

olya-2409 [2.1K]

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\quad 
\begin{cases}
r=radius\\
h=height\\
-----\\
r=1.5\\
h=18
\end{cases}\implies V=\pi \cdot 1.5^2\cdot 18$

3 0

4 years ago

The table on the left and the equation below represent

olasank [31]

Answer:

There isn't enough information

Step-by-step explanation:

In the table, on the side with a we can see that there is not one with 6 as a possible term. Therefore, we cannot find the term for b or c since you need both a and c to find b.

3 0

2 years ago

Estimate with one digit 6/253

aleksandr82 [10.1K]

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6 0

3 years ago

Can someone explain to me how to do this and give me the answer please?

pantera1 [17]

Answer:

To find x, y, z, p and q, you need to know about the property of Rhombus.

13y + 10 = 7y + 16 (pair of equal opposite sides)

=> 6y = 6

=> y = 1

x = 7x + 16 (2 equal adjacent sides)

= 7*1 + 16

=23

z = 90 deg (angle between 2 diagonals)

q = 22 deg (isosceles triangle)

p = 90 - 22 = 68 deg (right triangle)

4 0

3 years 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