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2 years ago

If these numbers represent elevations in meters, which is the farthest away from sea level?

Mathematics

1 answer:

Crank2 years ago

4 0

Answer:

The farthest away is 7. ( im not sure though ) sorry if its wrong :(

Step-by-step explanation:

I think its 7 because sea level is at 0 and 7 is the farthest away. You may be thinking that it could be one of the negatives but those aren't it because those aren't as far away as the 7. I AM SO SORRY IF ITS WORNG and please maybe give me brainliest because ive never had it. Please tell me if its worng.

You might be interested in

What is the midpoint of this segment??

Korvikt [17]

Answer:

A) 6,3

Why?

The midpoint, which is somewhat self-explanatory thanks to its name, is a point in the center of a line.

To answer this question properly, you need to analyze the line and assess the situation.

How did it go from x = 2 to x = 10? Let's see, if we count by two's across the line, you'll notice it makes more sense. So now we know that every square goes up by two. x = 6 is the exact center of the line.

I hope this wasn't confusing, it's tough to explain it!

3 0

3 years ago

Read 2 more answers

What numbers go in the sqaures?

givi [52]

On the first two pairs you need to write 3 and 4, and 6 and 2.

On the second page we have:

1) 4

2) 2 and 7

3) 1

4) 1 and 3

5) 3.

<h3>What numbers go in the squares?</h3>

a) Here we just need to use scientific notation.

For the first case, the numbers that go in the exponent boxes are the same ones that go in the example, which are 4 in the first one and 2 on the second one.

Then, if we want the difference to be 29,400, then we can use 3 on the first box and 6 on the second. OR, because this is an absolute value, we can just invert those.

This is:

6 and 2 on the first two squares, and 3 and 4 on the second two squares.

b) Here we want to make perfect squares.

For example, on the first case we have:

x*2*2

We need to find the value of x such that this is a perfect square. Notice that we can rewrite this as:

x*2*2 = x*4

Then it only is a perfect square if x = 4.

For the second case:

14*x*y

This is only a perfect square if x*y = 14

Then we can use x = 2 and y = 7.

With similar reasoning, on the next box we have:

2*8*x

2*8 = 16, is already a perfect square, so x = 1.

The next one is:

x*3*y

Here we can define x = 3 and y = 1, so we get a perfect square.

Finally:

3*x

Here the only possible value of x is 3.

If you want to learn more about perfect squares:

brainly.com/question/1538726

#SPJ1

5 0

2 years ago

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$