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Rzqust [24]
2 years ago
8

Let n = 14. What is the value of y in the equation y = 2 × n + 3?

Mathematics
1 answer:
Ghella [55]2 years ago
5 0

n=14 \\y=2\times n +3=2\times14+3 \\y=24+3\Longrightarrow y=\boxed{27}

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The area of the parallelogram below is square meters?
horsena [70]

Answer:

8 Square meters

Step-by-step explanation:

plus the 6m and 2m this meters is both side , so add that then it'll become 8 Square meters

or multiply the 6m and 2m it'll become 12m

3 0
2 years ago
Think of a number, divide it by 5 and then add 27 to the result. The resulting number is half of the original number. What was t
Salsk061 [2.6K]
This Is One Answer, There Are More Then One Answer.



100 / 5 = 20 + 27 = 47 / 2 = 23.5

100 = my number
5 = divide by

100 / 5 = 20
27 = add
20 = results of last answer
Then 20 + 27 = 47
47 = results of last answer
2 = half of
Then 47 / 2 = 23.5
7 0
3 years ago
Can someone please help with this? I'll give brainliest. I don't understand this... or the explanation it gives... the page befo
tresset_1 [31]

Step-by-step explanation:

Hello!
I'd love to help you learn.
I see the formula attached is 4*(\frac{1}{2})^{x} =-2^{x}-1\\

This problem is explaining you to graph, since algebraically it's a little more harder to try and find the solution.

Since they're both equal to each other, we can assign a variable like y, so we can make them into two individual lines.

You can use your scientific calculator to graph lines like this, or otherwise there are online sources (Desmos helps alot!) which can graph two equations as so.

Now that we have the corresponding system of equations, we can graph both.

y=4*(\frac{1}{2})^x\\ y=-2^x-1
Let's graph them on Desmos on the same plane. The answers are attached. Red is the first equation, blue is the second.

What determines the solutions of a system of equations?

-No solution: the two lines will never intersect/does not intersect ever, which means that there are no set point where it satisfies both equations.

-One solution: the two lines intersect once and only once, meaning that there is that one set point where the x and y values both satisfy both equations.

-Multiple solutions: the two lines will intersect each other multiple times, meaning there are multiple set points where the x and y values satisfy both equations. You usually will not have to worry about these problem sets.

-Infinite solutions: the two lines are both the same line, which means every x and y value will satisfy both equations.

Looking at the solution attached, we can see that there are no places where a system of equations intersect, therefore ruling it that they have no solution.

And to answer your last question, an asymptote is a imaginary line in which a equation can approach closely and closely, but will never touch that imaginary line. Think of a line at y=\frac{1}{x}. When you graph it, you can see that the two lines never ever EVER intersect the y-axis, or x=0; giving that x=0 as a vertical asymptote.

5 0
2 years ago
Which number best represents a helicopter ascending 112 ft? A. −112 B. +112 C. 112+ D. 112−
vampirchik [111]

Answer:

B +112

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Solve the system of equations.
Xelga [282]

Answer:

  b.  x=1, y=2, z=3

Step-by-step explanation:

The system of equations ...

  • 3x +2y +z = 10
  • 9x -6y +z = 0
  • x -y -3z = -10

has solution (x, y, z) = (1, 2, 3) . . . . matches choice B.

_____

While it is convenient to solve this using a graphing calculator or web site, one can easily solve the system by hand.

Subtract the second equation from 3 times the first:

  3(3x +2y +z) -(9x -6y +z) = 3(10) -(0)

  12y + 2z = 30 . . . . simplify

Dividing this result by 2 gives ...

  6y +z = 15 . . . . . . [eq4]

Subtract 3 times the third equation from the first:

  (3x +2y +z) -3(x -y -3z) = (10) -3(-10)

  5y +10z = 40 . . . . simplify

  y + 2z = 8 . . . . . . . divide by 5 . . . . . [eq5]

The two equations [eq4] and [eq5] can be solved any of the ways you usually solve two equations in two variables. Here, we'll use the first equation to write an expression for z that we can substitute into the second equation.

  z = 15 -6y . . . . . subtract 6y from [eq4]

  y + 2(15 -6y) = 8 . . . . . substitute for z in [eq5]

  -11y +30 = 8 . . . . . simplify

  -11y = -22 . . . . . . . subtract 30

  y = 2 . . . . . . . . . . . divide by the coefficient of y

  z = 15 -6(2) = 3 . . . . substitute for y in our equation for z

Substituting these values for y and z into the third original equation gives ...

  x - 2 -3(3) = -10

  x -11 = -10 . . . . . . . . simplify

  x = 1 . . . . . . . . . . . . add 11

The solution to the above system of equations is (x, y, z) = (1, 2, 3).

_____

<em>Comment on the problem statement</em>

Math is generally unforgiving of imprecision. The given system of equations has no variable "z", and some other typos are apparently involved. That is why we rewrote the system to the equations shown above.

It is very easy to mistake z for 2, or g for 9, or o for 0, or 1 for 7. There are other confusions that are possible, as well. Letters I (eye) and l (ell) are easily confused, and may be confused with 1 (one) as well. Sometimes y and 4, or 4 and 9, can also be written so as to be difficult to tell apart. Great care must be taken when handwriting these symbols.

7 0
3 years ago
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