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2 years ago

Deandre knit a scarf for each of his 8 friends. Each scarf was 57.6in

Mathematics

1 answer:

madreJ [45]2 years ago

7 0

Multiply 57.6 x 8 = 460. 8 and since you have to convert it to feet you divide by 12 which gets you 38.4 feet

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Ari thinks the perfect milkshake has 3 ounces of caramel for every 5

tangare [24]

Answer:

A - They Have Too Much Caramel

Step-by-step explanation:

Ari likes 3 oz caramel for 5 scoops ice cream.

Freeze Zone makes 6 oz caramel with 8 scoops of ice cream.

Divide both amounts of Freeze Zone by 2.

6 oz caramel for 8 scoops of ice cream is the same ratio as

3 oz caramel for 4 scoops of ice cream.

Since Ari likes 3 oz caramel for 5 scoops ice cream,

he will think it's too much caramel for the ice cream.

Answer: A - They Have Too Much Caramel

8 0

3 years ago

Write the fve-number summary for each set of data

oksian1 [2.3K]

Answer:

Step-by-step explanation:

first arrange the data from ascending to decending order

1,5,8,10,11,14,17,19,20,23

Q1=(N+1)/4

=(10+1)/4

=11/4

=2.75

=2 nd item+(3 rd -2 nd)term

=5+(8-5)

=5+3

Q2=(N+1)/2

=(10+1)/2

=11/2

=5.5 term

=(5 th term+6 th term)

=(11+14)/2

=25/2

=12.5

Q3=3(N+1)/4

=3(10+1)/4

=3*11/4

=33/4

=8.25

=8 th term +(9 th -8 th )term

=19+(20-19)

=19+1

=20

3 0

2 years ago

What’s 2 + 2? <br> A. 0<br> B. 1<br> C. 2<br> D. 4

snow_tiger [21]

Answer:

are u kidding me it's literally D --_--

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find the number of DEGREES rotated in a circle of radius 6 inches if an arc is created with a length of 11(pi) inches

artcher [175]

Radius = 6 inches

Arc lenght = 11pi inches

Degrees ?

Arc lenght = radius x angle in radians

A = r * a

Solving for a:

a = A/r = 11pi/6 = 5.759586532 radians

5.759586532 to degrees ==> a = 330 degrees

Answer:

330 degrees

3 0

8 months 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