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3 years ago

What is the area, in square centimeters, of the figure below? *

Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

Hello I believe the answer is 6.96 But I dont know for sure so answer that at your OWN risk.

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What is the slop of y=1x+3

vaieri [72.5K]

Answer:

the slope is:1 the y-intercept is:3

Step-by-step explanation:

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2 years ago

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I really need someone to answer this, QUICK!

Elodia [21]

Answer:

A. LCM=72 B. GCF=7 C. 7(5+9)

Step-by-step explanation:

A. 8/2=4/2=2 9/3=3 Use the denominators to find the LCM. In this case, the LCM is 72.

B. 35/5=7 63/3=21/3=7 The GCF for these numbers is 7 because it is the only factor that they have in common.

C. 7 x 14=88 Since 5 and 9 don't have common factors, they work perfectly for this problem.

6 0

3 years ago

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IceJOKER [234]

Answer: 3/4 x 48
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Got 12 by reducing the numbers with the Greatest Common Factor (GCF) 4

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3 years ago

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annie adds 2 spoonfuls of sugar to her tea and sophie adds 3 spoonfuls to her tea. If one spoonful is 4 grams of sugar and a a c

Tresset [83]

240%

Step-by-step explanation:

if there is 240 grams then you just put a percent sing but are you asking What is the percentage of sugar in each cup? by spoons?

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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