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3 years ago

HELP PLEASE :(( thank u

Mathematics

1 answer:

yaroslaw [1]3 years ago

8 0

Answer:

the 3rd one hope that helped

Step-by-step explanation:

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A number hidden by spilt ink multiplied by 12 equals sixteen thousand, eight hundred and thirty-six. Sixteen thousand, eight hun

Nezavi [6.7K]

Answer:

Step-by-step explanation:

Assume the hidden number is x;

So, if x*12= 16,836

Next, divide both sides by 12 to solve for x;

x= 16,836 / 12

x= 1,403

So the number hidden by spilt ink is 1,403.

Next, to answer the question, divide 16,836 by the hidden number,1403;

16,836 / 1403 = 12

Therefore, it will be equal to 12 which is basically the reverse of the first equation.

6 0

4 years ago

Help me please ASAP tysm

snow_lady [41]

Answer:

2/3 ft

Step-by-step explanation:

area of triangle =base*height/2

2/5=base*6/5/2

2/5 *2 =base * 6/5

4/5 =base*6/5

4/5÷6/5=base

4/5 * 5/6=base

20/30=base

2/3=base

therefore base=2/3 ft

3 0

3 years ago

What is the probability of spinning two on a spinner that has 5 sides?

k0ka [10]

1 in every 5 spins...
Or just a 20% chance

8 0

3 years ago

Select the correct answer.

Scilla [17]

The matrix equation that could show the total books is $\left[\begin{array}{cc}7&1\\2&6\end{array}\right] + \left[\begin{array}{cc}4&3\\6&1\end{array}\right] = \left[\begin{array}{cc}11&4\\8&7\end{array}\right]$

<h3>How to determine the matrix equation?</h3>

The given parameters are the two-way tables attached to the question.

From the tables, we have the following matrices

$June = \left[\begin{array}{cc}7&1\\2&6\end{array}\right]$

$July = \left[\begin{array}{cc}4&3\\6&1\end{array}\right]$

The sum of these matrices is represented as:

June + July = Total

So, we have:

$\left[\begin{array}{cc}7&1\\2&6\end{array}\right] + \left[\begin{array}{cc}4&3\\6&1\end{array}\right]$

Add the elements at corresponding positions

$\left[\begin{array}{cc}11&4\\8&7\end{array}\right]$

Hence, the matrix equation is $\left[\begin{array}{cc}7&1\\2&6\end{array}\right] + \left[\begin{array}{cc}4&3\\6&1\end{array}\right] = \left[\begin{array}{cc}11&4\\8&7\end{array}\right]$