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2 years ago

How do I do this number 5 and 6

Mathematics

1 answer:

grin007 [14]2 years ago

4 0

Answer:

5. 40

Step-by-step explanation:

(14) +(4x+6) = 8x

14 + 4x+ 6 = 8x

20 = 8x - 4x

4x = 20

x = 5

RT = 8x

= 8 * 5....since x = 5

= 40

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Is 3.667 a rational number,interger,or whole number

steposvetlana [31]

Rational 3667/1000
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3 years ago

Please help 2х -3y =18

siniylev [52]

Answer:

x = 3(6 + y)/2

Step-by-step explanation:

Solving for x

Add 3y to both sides.

2x = 18 + 3y

Divide both sides by 2.

x = 18 + 3y/2

Factor out the common term 3.

x = 3(6 + y)/2

3 0

2 years ago

What is the distance between points F(2, 9) and G(4, 14)? Round to the nearest whole number.

Fittoniya [83]

Answer:

Step-by-step explanation:

5 0

3 years ago

Solve each equation. Show your work<br><br> 13a=-5<br> 12−b=12.5<br> -0.1=-10c

amid [387]

1) 13a=-5
Make a the subject of the formula by dividing both sides by 13(the coefficient of a)
13a/13=-5/13
Therefore a= -0.385

The second one). 12-b= 12.5
You take the 12 to the other side making b subject of the formula (-b in this case)
-b= 12.5-12
-b= 0.5
(You cannot leave b with a negative sign so you will divide both sides by -1 to cancel out the negative sign)

-b/-1= 0.5/-1
Therefore b=-0.5

The third one). -0.1= -10c
You will divide both sides by the coefficient of c(number next to c) which is -10
-0.1/-10= -10c/-10

Hence, c= 0.01

5 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