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2 years ago

7

Marta uses 1 peice of paper and 1 ribbon to make a kites.the papre comes in 3 packs and the ribbon comes in 4 packs.what is the

least number of kites marta can make without any supplies left over

1 answer:

lianna [129]2 years ago

5 0

If I'm understanding this correctly then 3 is your answer, but to be sure you'd have to be specific on how many papers and ribbons are in each pack.

You might be interested in

a square has a side length of x. A rectangle has a length that is 4 inches longer than the square and a width that is 2 inches s

vodomira [7]

Answer:

Length of the rectangle is = 8 inches.

Step-by-step explanation:

Given :

Area of square = Area of the rectangle

According to the question:

Length of the square be 'x'.

Its area = (Side)*(Side) = $x^2$ ...equation (i)

Length of the rectangle = $(x+4)$

Width of the rectangle = $(x-2)$

Area of the rectangle =Length * Width

⇒ $(x+4)(x-2)=x^2-2x+4x-8$

⇒ $x^2+2x-8$ ...equation (ii)

Equating both the equation as area of both the figure are equal.

⇒ $x^2+2x-8=x^2$

⇒ $2x-8=0$ ...subtracting $x^2$ both sides

⇒ $2x=8$ ...dividing both sides with 2

⇒ $x=\frac{8}{2}=4$ inches

Plugging the value of x=4 in the length of the rectangle.

We have,

⇒ $(x+4)=(4+4)=8$

So the length of the rectangle = 8 inches.

4 0

3 years ago

Write the opposite of – 1/2 explain.

mrs_skeptik [129]

The opposite of -1/2 is 1/2. If you put -1/2 in absolute value, then the answer is 1/2.

8 0

3 years ago

Let P(n) be the statement that n! < nn where n is an integer greater than 1.

Ira Lisetskai [31]

Answer: See the step by step explanation.

Step-by-step explanation:

a) First, Let P(n) be the statement that n! < n^n

where n ≥ 2 is an integer (This is because we want the statement of P(2).

In this case the statement would be (n = 2): P(2) = 2! < 2^2

b) Now to prove this, let's complet the basis step:

We know that 2! = 2 * 1 = 2

and 2^2 = 2 * 2 = 4

Therefore: 2 < 4

c) For this part, we'll say that the inductive hypothesis would be assuming that k! < k^k for some k ≥ 1

d) In this part, the only thing we need to know or prove is to show that P(k+1) is also true, given the inductive hypothesis in part c.

e) To prove that P(k+1) is true, let's solve the inductive hypothesis of k! < k^k:

(k + 1)! = (k + 1)k!

(k + 1)k! < (k + 1)^k < (k + 1)(k + 1)^k

Since k < k+1 we have:

= (k + 1)^k+1

f) Finally, as the base and inductive steps are completed, the inequality is true for any integer for any n ≥ 1. If we had shown P(4)

as our basis step, then the inequality would only be proven for n ≥ 4.

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

A palindrome is a number that reads the same forward and backward. For example, 2442 and 111 are palindromes. If 5-digit palindr

Verdich [7]

Answer:

27

Step-by-step explanation:

Let the 5 digit palindrome formed from 1,2,3,4,5 be represented as XYZXY

Possible outcomes of X are 1,2,3 which give 3 possible outcomes

Possible outcomes of Y are 1,2,3 which also gives 3 possible outcomes

Possible outcomes of Z are 1,2,3 which will give 3 possible outcomes.

The total possible outcome = 3*3*3

= 27

6 0

3 years ago