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2 years ago

DE=3x,EF=x+1,and DF=1

Mathematics

1 answer:

liraira [26]2 years ago

8 0

Answer: 2.5

Step-by-step explanation:

$3x+x+1=11\\\\4x+1=11\\\\4x=10\\\\x=2.5$

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A campus club held a bake sale as a fund raiser, selling coffee, muffins, and bacon and eggs sandwiches. The club members charge

OLEGan [10]

Answer:

The number of bacon and egg sandwiches sells are 2 .

Step-by-step explanation:

Let us assume that the number of coffee cups campus club sells be x .

Let us assume that the number of muffins cups campus club sells be y .

Let us assume that the number of bacon and eggs sandwiches campus club sells be z .

As given

Campus club sold a total of 55 items .

Thus the equation becomes

x + y + z = 55

Of the club members sold 5 more muffins than cups of coffee .

Thus

Number of muffins = 5 + Number of cups of coffee

y = 5 + x

Putting the value of y in the equation x + y + z = 55 .

x + 5 + x + z = 55

2x + z = 55- 5

2x + z = 50

As given

The club members charged $1 for a cup of coffee, 3$ for a muffin , and $4 for a bacon and egg sandwich and earns $119 .

Thus the equation becomes

x + 3y + 4z = 119

Putting the value of y in the above

x + 3 × (x + 5) + 4z = 119

x + 3x + 15 + 4z = 119

4x + 4z = 119 - 15

4x + 4z = 104

Simplify the above

x + z = 26

Thus the two equations are

2x + z = 50

x + z = 26

Multiply x + z = 26 subtracted from 2x + z = 50.

2x -x + z - z = 50 - 26

x = 24

Put the value of x in x + z = 26 .

24 + z =26

z = 26 - 24

z = 2

Therefore the number of bacon and egg sandwiches sells are 2 .

7 0

4 years ago

Solve for x. −12x+13>35 Drag and drop a number or symbol into each box to correctly complete the solution.

Katen [24]

Answer:

x < -11/6.

Step-by-step explanation:

−12x + 13 > 35

-12x > 35 - 13

-12x > 22

Divide both sides by -12 and invert the inequality sign:

x < -22/12

x < -11/6.

7 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

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

$(x^2-(-4))+(y^2-(-4))=4$

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0

3 years ago

Does bin width matter when it comes to histograms? Why or why not?

maw [93]

Answer:

Avoid histograms with small bin widths that group data into lots of bins. A histogram constructed with small bin widths will show the distribution as a “pancake.” so no it does not help us see the pattern in the data.

Step-by-step explanation:

4 0

3 years ago