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Vesna [10]
2 years ago
8

There are between 90 and 115 cars in a parking lot. Exactly one-fourth of them are green. Exactly one-eighth of them have a stic

ker on the back that says, Mr. Donovan is the best teacher ever. How many cars MIGHT be in the parking lot
Mathematics
1 answer:
Schach [20]2 years ago
5 0

There are 104 cars in the parking lot.

According to statement there are between 90 and 115 cars on the lot.

So, {X| 90 < x < 115} (This renders an infinite solution set finite)

AND exactly one eight of them have a sticker on the back, so the total number of cars must be evenly divisible by eight.

X ∈ {96, 104, 112,}

AND exactly one fourth of the cars are green, so the number of cars must be evenly divisible by 4. Here all above written numbers are divisible by 4. So, find the mean to calculate the number of cars in the parking lot.

x = (96+104+112)/3

x = 104

There are 104 cars in the parking lot.

Learn more about ELIMINATION METHOD here brainly.com/question/13729904

#SPJ4

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Perform the indicated operation and simplify the result.
Finger [1]

Answer:

7/ (3a-1)

Step-by-step explanation:

3a^2 -13a +4        28+7a

------------------ * --------------------

9a^2 -6a+1         a^2 -16

Factor

(3a-1)(a-4)           7(4+a)

------------------ * --------------------

(3a-1) (3a-1)        (a-4)(a+4)

Cancel like terms

1                    7

------------------ * --------------------

(3a-1)                  1

Leaving

7/ (3a-1)

3 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
graph the image of this figure after a dilation with a scale factor of 2 centered at the origin use the polygon tool to graph t
Iteru [2.4K]

Answer:

For this answer, I will label the points. Starting at the top left, then top right, then bottom left and bottom right let the points be A, B, C, D.

The new coordinates will be

A(-4,10)

B(4,10)

C(-4,4)

D(4,4)

Step-by-step explanation:

The question is asking for a dilation which is a transformation that makes an image proportionately smaller or larger by a scale factor. The scale factor is how much smaller or larger the shape will be, if the scale factor is between 0 and 1 then it will shrink, if it is greater than one then the image will stretch (be larger). In this case, the scale factor is 2, therefore the image will stretch. Since the center of dilation is the origin, to find the new coordinates simply multiply each x and y value by the scale factor. So A's original coordinates (-2,5) become (-4,10) and so forth. Therefore the equation for this dilation is (x, y) → (2x,2y).

4 0
3 years ago
Read 2 more answers
I need help with this please :)
Marina CMI [18]

9514 1404 393

Answer:

  all are Not Equivalent

Step-by-step explanation:

The coefficient of t³ in the target expression is -3. In order for that to be the result of simplifying any of the given expressions, they must have ( )^(t^3) in the denominator. None do, so none of the offered choices is equivalent to the given expression.

_____

The expressions simplify like this:

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4 0
3 years ago
The circumference of the target above is 6,537.48 mm.
malfutka [58]
2,082 mm







Hope this helps
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3 years ago
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