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3 years ago

10

A rectangle has a length of 1.5 feet and width of 3inches. What is the area of a rectangle in square inches?what is the area in

square feet? What is the area of the rectangle in square inches? Simplify answer

1 answer:

grigory [225]3 years ago

7 0

The answer would be 54 square inches and .375 square feet.

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12. How many students scored at least 60 on the test?<br> a) 3<br> b) 4<br> c) 11<br> d) 7

Troyanec [42]

4
i don’t have an explanation

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4 years ago

Mr. Cooper's garden is 28 feet long and 4 feet wide.<br> What is the area of his garden?

Evgesh-ka [11]

Answer:

112 feet squared

Step-by-step explanation:

Just multiply the length and the width then you get the area which is 112 ft squared.

8 0

3 years ago

Read 2 more answers

Line p goes through points (-2,6) and (4,-3) what is the equation of line p?

coldgirl [10]

Answer:

-3/2x+3

Step-by-step explanation:

y=mx+c

y=-3/2x+c

Substitute x and y values,

c=3

Therefore,

equation of line p=-3/2x+3

3 0

3 years ago

15 and 25 least common multiple

dsp73

If you want to add or subtract fractions :

1- make sure that the denominators of the fractions are equal. if they are not, then multiply each of them by a number so that they will be equal.

2- add or subtract the numerators.(if you already had multiplied the denominator of a fraction by a number, then you should multiply the numerator by the same number and do the adding or subtracting)

3- simplify it's possible

here are the answers :

$\frac{3}{7} + \frac{6}{11} = \frac{33 + 42}{77} = \frac{75}{77}$

$\frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12} = 1 \frac{1}{12}$

$\frac{4}{5} + \frac{3}{7} = \frac{28 + 15}{35} = \frac{43}{35} = 1 \frac{8}{35}$

$\frac{7}{12} - \frac{1}{8} = \frac{14 - 3}{24} = \frac{11}{24}$

$\frac{9}{10} - \frac{2}{3} = \frac{27 - 20}{30} = \frac{7}{30}$

good luck

5 0

4 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:

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

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

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

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

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

$\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)$

7 0

2 years ago