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3 years ago

Unit 1: Functions

Mathematics

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

hey mate I don't know it's answer

Step-by-step explanation:

sorry!

do your homework at your own

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Andrew, who operates a laundry business, incurred the following expenses during the year. ∙ Parking ticket of $250 for one of hi

marysya [2.9K]

Answer:

Option d: $0

Step-by-step explanation:

Parking penalties arent deductible from expenses, so Andrew cant deduct neither the 250 for his delivery van nor the 75 for the rock concert. Dui tickets definetely arent deductable from expenses, so Andrew cant deduct the 500 from the ticket either. That gives us only 2 options: Andrew can deduct $600 (if he is able to deduct from attoeney's fee), else he cant deduct anything.

However, the attorney's fee shoudn't be deductible, because as a general rule, personal legal fees are not deductible. So the answer is option d, $0

7 0

3 years ago

I got more coming btw.

fiasKO [112]

Answer:

The answer is C

Step-by-step explanation:

None

6 0

3 years ago

Read 2 more answers

Mary can build 2 sand castles ( c ) in 4 hours ( t ). Write an equation in the form of c=rt that represents the relationship bet

wariber [46]

2 = r(4)
To find r, you've gotta divide it out.
2/4 = r/4
1/2 or 0.5 = r

c=(0.5)t

8 0

4 years ago

Select the correct answer. Which set of vertices forms a parallelogram? OA A2,4), B(3, 3), C6, 4), D(5, 6) OB. A(-1, 1), B(2, 2)

Tanya [424]

Answer:

(6,4)

Step-by-step explanation:

^^^

4 0

2 years ago

Solve the simultaneous equations<br> y = 9 - X<br> 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

4 years ago