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3 years ago

What is 3 1/4 minus 1 1/3?

Mathematics

2 answers:

natita [175]3 years ago

8 0

Answer:

1 11/12

Step-by-step explanation:

Simplify 3\times 43×4 to 1212.

\frac{12+1}{4}-1\frac{1}{3}

12+1

−1

3 Simplify 12+112+1 to 1313.

\frac{13}{4}-1\frac{1}{3}

−1

4 Convert 1\frac{1}{3}1

to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

5 Simplify 1\times 31×3 to 33.

13/4 -3 +1 / 3

6 Simplify 3+13+1 to 44.

13/4-4/3

7 (Find the Least Common Denominator (LCD) of 13/4,4/3 In other words, find the Least Common Multiple (LCM) of 4,34,3.)

LCD = 1212

8 (Make the denominators the same as the LCD)

13 x 3 - 4 x 4

4 x 3 -3 x4

9 (Simplify. Denominators are now the same.)

39/12-16/12

10 (Join the denominators.)

39-16

39−16

11 Simplify.

23/12

12 (Convert to mixed fraction.)

1 11/12

MissTica3 years ago

7 0

Answer:

23/12 or 1 11/12

Step-by-step explanation:

13/4 - 4/3

39/12 - 16/12

23/12

1 11/12

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Among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a ma

scoundrel [369]

Answer:

Step-by-step explanation:

Given that among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course.

From the above we find that

a) either economics of Math course is

$315+213-123 =405$

Out of 500 students 405 have taken either Math or Economics

Hence

c) student who have taken neither = $500-405 =95$

Exactly one course is either math or economics - both

= $405-123 = 282$

3 0

3 years ago

Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3

nlexa [21]

Answer:

$(x,y) = (5,-6)$

Step-by-step explanation:

$\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\$

$~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)$

$\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\$

$\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5$

$y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6$