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3 years ago

which of the following best describes the english expression five more thats the quantity of a number minus six

Mathematics

2 answers:

Alborosie3 years ago

4 0

Answer:

n,,n,n,n

Step-by-step explanation:

kupik [55]3 years ago

3 0

Answer:

5+(x-6)

Step-by-step explanation:

Expression: 5 more than the quantity of a number minus 6

Minus indicates subtraction.

5 more than the quantity of a number - 6

Quantity is used to describe an unknown number.

Let that quantity be x.

More than indicates addition (+)

5+(x-6)

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Need help finding out something, refer to picture

Alja [10]

Answer:

He made a mistake in the second line.

Step-by-step explanation:

He factored out x out of 6-2x and 5+4x, which is not possible because there is not an x in both 6 and 5.

8 0

3 years ago

Read 2 more answers

Which expression is NOT equivalent to −38⋅(−4+12)? Drag this expression to the box.

lesya [120]

I had this question so here's the answer and explanation! :)

Answer:

The answer to your question is the first option :D

Step-by-step explanation:

Original expression

-3/8 (-4 + 1/2)

First option -3/8 (-4) + (3/8)(1/2) This option is not equivalent because

they forgot the negative sign of the

second term.

Second option (-3/8)(-4) + (-3/8)(1/2) This option is equivalent to the

original. Distributive property

Third option (-3/2)(-3 1/2) This option is equivalent to the original

Fourth option (-3/8)(-3) + (-3/8)(-1/2) This option is equivalent to the original

HOPE THIS HELPS :)

5 0

3 years ago

Read 2 more answers

Please solve this question faaaast

Kazeer [188]

Answer:

300-400

Step-by-step explanation:

The first step is finding the total of the data we have. So, we take 5 + 10 + 15 + 20 + 25 + 15 + 10 which equals 80.

The median is the middle point of all the data. If it's an odd number, you can calculate the median with the equation (n+1) / 2, plugging in the total amount of data for n.

When it's an even number though, there is no one middle point since the data splits evenly in 2, so we have to use 2 equations: n/2 & (n/2) + 1. Then, we find the average of those two data points. (Although, you don't have to do that for this particular question).

When we plug 80 in for n in both of these equations, we get 40 and 41.

To where this is in the question, we have to count up from the bottom of the chart. 1-5 is below 100, 6-15 is 100-200, 16-30 is 200-300, and 31-50 is 300-400.

Since 40 and 41 are between 31 and 50, the answer is 300-400.

Hope this helps! :)

6 0

2 years ago

Rosetta has 1/8 bag of cat food left. She divides the remaining of the cat food into 12 equal sized portions. What fraction of a

Nookie1986 [14]

Answer:

1/96

Step-by-step explanation:

6 0

3 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

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

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

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago