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3 years ago

I need help on this really bad i don't know how to do it.

Mathematics

2 answers:

Roman55 [17]3 years ago

8 0

Answer:

x= positive 5 y= negative 5

Step-by-step explanation:

i know this because that is were the line is crossing have a nice day :))

Marizza181 [45]3 years ago

5 0

X equals positive 5 and Y equals negative 5. I hope you have a wonderful day

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If you solve this correctly I’ll give you brainliest!

Evgesh-ka [11]

I would consider the number 2

3 0

2 years ago

A bill from a plumber was $306 The plumber charged $150 for parts and $52 per hour for labor. How long did ths plumber work at h

ludmilkaskok [199]

We can make an equation from this.
Our intial value is 150 for parts, and he works for 52 dollars an hour.
We can define h for hours and c for total cost.
Therefore, we have:
c = 150 + 52h.
We can set this equal to 306 dollars, as that's the given total cost:
306 = 150 + 52h
Subtract 150 from both sides:
52h = 156
Divide both sides by 2:
h = 3
He worked 3 hours.
hope this helped!

6 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

1 year ago

A standard brick is 3.625 inches wide, 2.25 inches high, and 7.625 inches long. How many bricks could be in a pallet of bricks t

Snezhnost [94]

Answer: Option D. 1,778

Solution:

Standard brick:

Width: w=3.625 in

Height: h=2.25 in

Length: l=7.625 in

Volumen of one standard brick: v

v=w*h*l

v=(3.625 in)*(2.25 in)*(7.625 in)

v=62.19140625 in^3

Pallet of bricks:

Side: s=4 feet

s=(4 feet)*(12 in / 1 feet)→s=48 in

Volume of a pallet of bricks: V=s^3

V=(48 in)^3

V=110,592 in^3

Number of bricks could be in a pallet: n

n=V/v

n=(110,592 in^3) / (62.19140625 in^3)

n=1,778.252119

n=1,778

6 0

3 years ago

Read 2 more answers

Hi! What is y = x^2 -4x +5 vertex form? Please explain how you got this answer. Thanks! :)

Anarel [89]

${x}^{2} - 4x + 5 = 0 \\ {x}^{2} - 4x = - 5 \\ \\ { (\frac{b}{2} )}^{2} = ( \frac{ - 4}{2} ) = ( { - 2})^{2} = 4 \\ {x}^{2} - 4x + 4 = - 5 + 4 \\ {(x - 2)}^{2} = - 1 \\ {(x - 2)}^{2} + 1$
The last line is in vertex form, and can see the vertex is located at (2,1).

8 0

2 years ago

Read 2 more answers