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kirill [66]
2 years ago
7

4. A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the

Mathematics
2 answers:
ankoles [38]2 years ago
7 0

Answer: the answer is simply 5.2cm

Step-by-step explanation:

serious [3.7K]2 years ago
5 0
Rather than memorize sines and cosines of various common triangles, I prefer to remember the simple derivation of those formulas…

Imagine that the wall is a mirror, so you have one ladder leaning against the wall and you also have the reflection of that ladder leaning behind it. Now the ladder is 6m long, its reflection is 6m long, and if the distance between their bases is also 6m then you have a 60–60–60 equilateral triangle. And, good news, the problem says that the base angle is 60 degrees. So the real part in front of the mirror is half of the equilateral triangle, which means the distance from the base to the mirror is 3m. And then the height is sqrt(36–9) = sqrt(27) = 3*sqrt(3).

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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
Can you answer those questions
alexandr402 [8]

Step-by-step explanation:

the image u have shared is too much covered by focus light send it again

7 0
3 years ago
I need to find the area of the shaded region
spin [16.1K]
Well first find the area of the semi-circle. 

If the area of a cricle is equal to pi*radius^2 , then you can just find that and divide by 2.

So, A = pi*2^2. = 4pi

We know that the radius is 2 because the length of the side of the rectangle is 4, meaning that the diameter of the semi-circle is 4, and so the radius is 2 as it is half of the diameter. 

We can easily calculate the area of the rectangle, which is
 Length * width = 6*4 = 24.

Next we divide 4pi by 2 in order to get the area of the semi-circle, giving us an area of 2pi

We can just subtract 2pi from 24 and get the area of the shaded region.

Area of the shaded region (answer): 17.7
4 0
3 years ago
An ordered pair of real numbers can be represented in a plane called the rectangular coordinate system or the ___________ plane
Ghella [55]
Coordinate plane is the answer
5 0
3 years ago
The average number of points a basketball team scored for
Ganezh [65]
Average is

(2x + X+6)/3=63
(2x+x+6)= 189
3x+6=189
3x=183 61

X=61
Scores for the game are: 61, 61, and 67
3 0
3 years ago
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