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3 years ago

Write a number sentence to add 6 ones to 21

Mathematics

2 answers:

makvit [3.9K]3 years ago

7 0

6 + 21 = 27 (Add six ones)
60 + 21 = 81 (Add six tens)

wolverine [178]3 years ago

7 0

Answer:

The required number sentence is 6 + 21 = 27.

Step-by-step explanation:

We are required,

To write a number statement in order to add 6 ones to 21.

Now as we know,

'Ones' refer to the unit number 1.

Then '6 ones' means that the number is 6×1 = 6

Thus, we get that,

Hence, the required number sentence is 6 + 21 = 27.

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Sarah cuts a piece of rope that is 4.5 feet long.

icang [17]

Answer:

4.5 feet = 1.5 yards

Step-by-step explanation:

1 yard equals 3 feet

Then:

1 ----- 3

x -----4.5

$x = 4.5/3 = 1.5 yards$

Hope this helps

3 0

2 years ago

The ratio x:10 is equilvalent to 9:6. What is the value of x

Troyanec [42]

Answer: x=20/3=6.67

Step-by-step explanation:

To solve proportions, the most useful way is to use [cross multiplication], where we not multiply horizontally but as diagonally.

-----------------------------------------------------------------------------

Given

x:10=9:6

Cross Multiply

9·x=6·10

9x=60

Divide 9 on both sides

9x/9=60/9

x=20/3=6.67

Hope this helps!! :)

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4 0

3 years ago

118 {121÷(11×11)-(-4)-(3-7)}

Rama09 [41]

Question:

To Simplify:

118 {121÷(11×11)-(-4)-(3-7)}

Solution:

↠118 {121÷121+4-(-4)}

↠118 {1+4+4}

↠118 {5+4}

↠118{9}

↠118×9

↠1062

▬▬▬▬▬▬▬▬▬▬▬▬

The upper Solution is done by applying BODMAS

About BODMAS:

B→ Bracket

O→ Of

D→ Division

M→ Multiplication

A→ Addition

S→ Subtraction

5 0

2 years ago

Read 2 more answers

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

Abc~def what sequence of transformation will move ABC onto def

DENIUS [597]

Answer:

Step-by-step explanation:

8 0

3 years ago