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1 year ago

A chemical balance needs to weigh 237g using weights of 50g, 20g, 10g and 1g. How many of each weight is needed?

Mathematics

2 answers:

Delicious77 [7]1 year ago

4 0

The required number of weights used in chemical balance to weigh 237g is 4,1,1,7 respectively.

What is mathematical expression?

A mathematical expression expression in mathematical is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

Here, it is given that :

A chemical balance needs to weigh 237g using weights of 50g, 20g, 10g and 1g.

Now, let us find the number of weights for each weight type :

For 50g weight the maximum number of weights needed will be 4 that is :

50g x 4 = 200g

Now, for 20g weight the maximum number of weights needed will be 1 that is :

20g x 1 = 20g

Now, for 10g weight the maximum number of weights needed will be 1 that is :

10g x 1 = 10g

Now, for 1g weight the maximum number of weights needed will be 7 that is :

1g x 7 = 7g

Adding all the weights we get :

200g + 20g + 10g + 7g = 237g;

which is the required weight and the number of weights used is 4,1,1,7 respectively.

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3 years ago

Read 2 more answers

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

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