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2 years ago

15

13 cars are parked in a parking lot. each car has four tires. if seven of the tires are damaged then him many tires are in the p

arking lot are not damaged?
please help me !!

2 answers:

myrzilka [38]2 years ago

7 0

Answer:

45

Step-by-step explanation:

13 * 4=52

52-7=45

45 tires not damaged

malfutka [58]2 years ago

4 0

Answer:

45

Step-by-step explanation:

13 x 4 = 52

13 represents # of cars

4 represents tires per car

52 represents total tires

52 tires - 7 damaged tires = 45 undamaged tires

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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

3 years ago

Which fraction is bigger 5/6 or 8/9?

dimulka [17.4K]

5/6 = 5 ÷ 6 = 0.8333

8/9 = 8 ÷ 9 = 0.8888

so 8/9 is bigger.

hope this helped, God bless!

4 0

3 years ago

Read 2 more answers

I need to find Domain and Range but how do I know when to use brackets or parentheses

german

Answer:

Square brackets [ or ] are used when the end point is included.
Parentheses ( or ) are used when the end point is not included.

Step-by-step explanation:

The domain is the horizontal extent of the graph. Here, it is from -2 to +3, with -2 included in the domain and +3 not included.

As an inequality, the domain is written ...

-2 ≤ x < 3

In interval notation, we use a square bracket for the "or equal to" case:

[-2, 3)

__

The range is the vertical extent of the function. Here, the range extends from -4 (not included) to +5 (included).

As an inequality, the range is written ...

-4 < y ≤ 5

In interval notation, the range is written using a square bracket for the "or equal to" case:

(-4, 5]

_____

<em>Comment on infinity</em>

There is no such thing as "equal to infinity" so an infinite limit is always represented in interval notation using a parenthesis.

3 0

2 years ago

Point P is the center of the circle below. What is the exact area

Ber [7]

Answer:

192π unit²

Step-by-step explanation:

Given that :

Radius = 24

Area of circle = πr²

Area = π*24²

Area = 576π in²

The Shaded area is 120°

Entire Circumference = 360°

Hence, shaded area = 120°/ 360° = 1/ 3 of the area

1/3 * 576π in²

= 192π unit²

5 0

2 years ago

Question in picture above

almond37 [142]

A is the correct answer

8 0

3 years ago