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raketka [301]
1 year ago
14

Stem and leaf plots And box Pls help I have no idea what this is

Mathematics
1 answer:
AlladinOne [14]1 year ago
6 0

Answer:

32

Step-by-step explanation:

The end of the box is at 32.

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What is 0.4 divided by 0.242? Divide and check your answer using multiplication. Show your work.
Setler79 [48]
0.4 divided by 0.242 = 0.605 then you can either multiply. 0.4 times 0.605 and get 0.242 or you can do 0.242 times 0.605 and get 0.4 either way will work
5 0
3 years ago
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The length of a rectangle is 2 times the width of the rectangle. Let x represent the width of the rectangle.
nikitadnepr [17]

Answer:

ok ASAP done ..... for it

4 0
3 years ago
Math<br><br><br><br> pls help!!<br><br><br><br><br><br> answers?
statuscvo [17]

Answer: Choice B) Infinitely many solutions

  • one solution: x = 8, y = -7/2, z = 0
  • another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: (-1/4)*R3 \to R3

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l}  \ \\\ \\(-1/4)*R3 \to R3\\\end{array}

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\\ \\R3-R1 \to R3\\\end{array}

Whenever you get an entire row of 0's, it <u>always</u> means there are infinitely many solutions.

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

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\5*R1-R2 \to R2\ \\\ \\\end{array}

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

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

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2

which simplify to

R1: x+2z = 8\\R2: y-z = -7/2

Let's get the z terms to each side like so:

x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\

Therefore, all of the solutions are of the form (x,y,z) = (-2z+8, z-7/2, z) where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just <em>any</em> triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

8 0
2 years ago
Evan has 2 pieces of wood: one is 90 inches and the other is 72 inches.he wants to cut both pieces of wood into smaller pieces s
Gre4nikov [31]

72/2=36

90/2=45

2+2=4 pieces of wood

6 0
2 years ago
I need to know how to solve <br> -4-(2+4(-6)-4-(2^2-5*2))
sweet-ann [11.9K]

Answer:

16

Step-by-step explanation:

This problem requires PEMDAS

Parentheses ( )

Exponents ^

Multiplication

Division

Add

Subtract

Start by solving anything in parentheses. There's an exponent within the parentheses, so we change that 2^2 into 4 and also make sure to multiply 5 x 2 before subtracting.

-4 - (2 + -24 - 4 - (4-10))

-4 - (2 + -24 - 4 - (-6))

Again, solve parentheses first.

-4 - (-22 - 4 - (-6))

-4 - (-26 + 6)

-4 - (-20)

-4 + 20

Answer is 16

4 0
3 years ago
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