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3 years ago

Solve each system by substitution. please help me 2x{+y=6 x{+y=3

Mathematics

1 answer:

saw5 [17]3 years ago

6 0

1. x= 3/Y as a fraction btw

2. Same thing

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If x to the 2nd power equal 60, What is the value of x

Sedaia [141]

Answer:

7.745

Step-by-step explanation:

Square root of 60 equals X.

5 0

3 years ago

Diameter = 18 cm

miskamm [114]

Answer: B. 254.34 cm^2

Step-by-step explanation:

The area of a circle is pi times the radius squared.

Using the diameter we have to find the radius.

The diameter is two times the radius meaning that the radius is half the diameter.

18/ 2 = 9

The radius will be 9 cm

Now using the formula v= pi * r^2 , input the value for pie and the radius and solve for v the volume

v = 3.14 * 9^2

v= 3.14 * 81

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7 0

2 years ago

Read 2 more answers

Find the measure of the missing value in the

kumpel [21]

Answer:

12.6

Step-by-step explanation:

The two left sides have lengths, so this allows you to establish the ratio of the lengths of the sides of the triangles,

left triangle : right triangle

ratio = 25 : 17.5

The bottom sides are also in the same ratio.

ratio = 18 : w

Write a proportion by setting the ratios equal and solve for w.

25/17.5 = 18/w

Cross multiply.

25w = 17.5 * 18

25w = 315

w = 12.6

Answer: 12.6

4 0

3 years ago

Solve 6 1/2 + 7 3/8 - 6 1/2

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

6 1/2 - 6 1/2 + 7 3/8 =

0 + 7 3/8 = 7 3/8

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:

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

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

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

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

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

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

7 0

2 years ago