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3 years ago

Find the missing side of the triangle. Then round to the near test tenth if necessary

Mathematics

1 answer:

zlopas [31]3 years ago

8 0

Answer:

I think it might be 78. I'm so sorry If I'm wrong I'm.not the bet at them but just try it

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Divide x² +x +1 by x+ 1 by long division method

fiasKO [112]

Answer:

remainder 1...........

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2 years ago

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Can y’all help me on question 19?!

m_a_m_a [10]

Answer: 178.31

Step-by-step explanation: subtract 574.54 by 396.23

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3 years ago

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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

Determine the values of the variables in isosceles trapezoid CHLE below.

Alex73 [517]

Answer:

The values of variables x and m are 11 and 17

Step-by-step explanation:

The question has missing details as the diagram of the trapezoid isn't attached.

(See attachment).

Given that trapezoid CHLE is isosceles then the angles at the base area equal (4x)

And

The angles at the top are also equal

8m = 11x + 15

At this point, the four angles in the trapezoid are 8m, 11x + 15, 4x and 4x..

The sum of interior= 360

So,

11x + 15 + 8m + 4x + 4x = 360

Collect like terms

11x + 4x + 4x + 8m = 360 - 15

19x + 8m = 345

Substitute 11x + 15 for 8m

19x + 11x + 15 = 345

30x + 15 = 345

30x = 345 - 15

30x = 330

Divide through by 30

30x/30 = 330/30

x = 11

Recall that 8m = 11x + 15;

8m = 11(11) + 15

8m = 121 + 15

8m = 136

Divide through by 8

8m/8 = 136/8

m = 17

Hence, the values of variables x and m are 11 and 17

7 0

3 years ago

A cone has volume 3. If the cone's radius is 1. what is its height?

Pani-rosa [81]

2.87 units

The volume of a cone is found by using the formula, V = (1/3)(pi)(h)r^2

Substituting values:

3 = (1/3)(3.14)(h)(1)

2.87 = h

6 0

3 years ago

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