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katrin2010 [14]
3 years ago
8

SOMEONE PLEASE HELP ME THIS IS DUE IN 30 min

Mathematics
1 answer:
Ksivusya [100]3 years ago
8 0

Answer:

its 560 meters because you do 10 times 14 which gives you 140 then you multiply 140 by 4 and you get 560

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At a certain time of the day, a tree 15 m tall casts a shadow of 12 m, while a second tree casts a shadow of 20 m. How tall is t
Anna35 [415]
We can write a proportion: 15 m is to 12 m as some other height h is to 20 m

So we get
\frac{15}{12}=\frac{h}{20}
\rightarrow 15*20=12h
\rightarrow 12h=300
\rightarrow h=25

So the second tree is 25 m tall.
7 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
3 years ago
25 is 36% of what number
ch4aika [34]

Answer:

x = 69.44

Step-by-step explanation:

We have, 36% × x = 25

or, 36/100 × x = 25

Multiplying both sides by 100 and dividing both sides by 36,

we have x = 25 × 100/36

x = 69.44

If you are using a calculator, simply enter 25×100÷36, which will give you the answer.

4 0
2 years ago
A child from a class of 30 students is chosen to pick one of five prizes. How many outcomes are there
Mamont248 [21]

Answer: 5?

Step-by-step explanation:

there are 5 prizes to be chosen from.

4 0
3 years ago
I NEED HELP PRONTO PLZZZ HELP ME
forsale [732]

Answer:

12 ÷ 4

Step-by-step explanation:

Let's try to solve the given expression to see if it matches with any of the options guven..

\frac{1}{4}*12

= \frac{1*12}{4}

= \frac{12}{4} (this is equivalent to 12 ÷ 4, which is given in the question)

Thus,

12 ÷ 4 equivalent to \frac{1}{4}*12

4 0
3 years ago
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