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Salsk061 [2.6K]
3 years ago
5

Jada has picked 1 cup of strawberries for a cake, which is enough for ¾ of the cake. How many cups does she need for the whole c

ake?
Mathematics
2 answers:
Maksim231197 [3]3 years ago
6 0

Answer:

she needs 1 \frac{1}{3}   cup  of strawberries

Step-by-step explanation:

<em><u /></em>

To find the number of cups needed, we will simply solve using proportion

From the question, Jade picked  1 cup of strawberries for 3/4 of a cake

and we are asked to find the number of cups needed for a whole cake

Let x be the number of cups needed for the whole cake

\frac{3}{4}   cake       =      1 cup of strawberries

1 cake     =       x

Cross multiply

\frac{3}{4}   ×  x     =    1  ×    1

\frac{3x}{4}    =   1

cross multiply

3x =4

To get the value of x, we simply divide both-side of the equation by 3

\frac{3x}{3}    =      \frac{4}{3}

(On the left-hand side of the equation, 3 will cancel out 3 leaving us with x and on the right-hand side of the equation 4 will be divided by 3 which will give us 1 \frac{1}{3} )

x =  1 \frac{1}{3}

Therefore  1 \frac{1}{3}  cup of strawberries is needed for the whole cake.

Stolb23 [73]3 years ago
5 0
She needs 1/4 more or .25cup more for the whole cake
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Answer:

a. False

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Step-by-step explanation:

Hi,

We have certain properties for matrices,<em> (where A and B are nxn matrices and I is the identity matrix) </em>:

{(A^{-1})}^{-1} = A

(AB)^{-1} = B^{-1}A^{-1}

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Using these properties, we verify the provided statements:

A. False.

None of the properties help verify this statement. We ca use an example for counter:

Let  A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right] and  B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right] , we calculate the L.H.S:

A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right]

The square of (A+B):

(A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right]

Lets calculate the R.H.S:

A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3)  &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right]

A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right]

This proves that: L.H.S ≠ R.H.S

Hence, A is false.

B. False

This can only hold when the eigenvalues for A are real.

trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}

C. True

This is a simplification of the distribution property of matrices.

D. True

The property that inverse is possible for any "n" value of the matrix.

E. False

Similar to part A, we can show that this property is invalid for any nxn matrix. Let:

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A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right]  = \left[\begin{array}{cc}2&0\\0&2\end{array}\right]

(A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right]

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Since, L.H.S ≠ R.H.S, the statement is false.

F. False

This is a basic matrix rule, that commutative property does not apply on matrices.

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