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2 years ago

Can Someone plz help me with this question? I will mark brainliest

Mathematics

1 answer:

Gala2k [10]2 years ago

8 0

The answer to your question is

No

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food ,treats ,toys ,bed, collar ,leash, Veterinary Care ,daily dog walking ,boarding ,and training classes which ones go under v

Aneli [31]

Variable Expenses: Food, Treats, Toys, Bed, Collar, Leash
Fixed Expenses: Veterinary Care, Daily Dog Walking, Boarding, Training Classes

4 0

3 years ago

Read 2 more answers

Solve the simultaneous equations y = 9 - X y = 2x2 + 4x + 6

kenny6666 [7]

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

Step-by-step explanation:

Given the simultaneous equations

$y=9-x$

$y\:=\:2x^2\:+\:4x\:+\:6$

Subtract the equations

$y=9-x$

$-$

$\underline{y=2x^2+4x+6}$

$y-y=9-x-\left(2x^2+4x+6\right)$

$\mathrm{Refine}$

$x\left(2x+5\right)=3$

$\mathrm{Solve\:}\:x\left(2x+5\right)=3$

$2x^2+5x=3$ ∵ $\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$2x^2+5x-3=3-3$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$=\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{4}$

$=\frac{-5+7}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Similarly,

$x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{2},\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

3 0

3 years ago

Assume f(x) = ax^2+ bx + c Which one is true? f is a parabola that opens sideways to the right. f is a parabola that opens sidew

Viefleur [7K]

Step-by-step explanation:

f is a parabola that opens upward.

5 0

2 years ago

Which equation represents the line through (4, -6) and (0, 2)?

lukranit [14]

Answer:

$y=-2x+2$

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when x=0).

1) Determine the slope (m)

$m=\displaystyle\frac{y_2-y_1}{x_2-x_1}$ where two points that fall on the line are $(x_1,y_1)$ and $(x_2,y_2)$

Plug in the given points (4, -6) and (0, 2):

$m=\displaystyle\frac{-6-2}{4-0}\\\\m=\displaystyle\frac{-8}{4}\\\\m=-2$

Therefore, the slope of the line is -2. Plug this into $y=mx+b$ :

$y=-2x+b$

2) Determine the y-intercept (b)

The y-intercept occurs when x=0. We are given that (0,2) falls on the line, so therefore, 2 is the y-intercept. Plug this into $y=-2x+b$ :

$y=-2x+2$

I hope this helps!

3 0

2 years ago

Read 2 more answers

There are 896 people at the school's basketball game. The

Setler [38]

Answer:

56 that's easy.....

Step-by-step explanation:

....

8 0

2 years ago