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2 years ago

I NEED HELP WITH MATHH

Mathematics

1 answer:

Nesterboy [21]2 years ago

6 0

Hey I can help give me one moment .

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At the beginning of its route, a garbage truck accelerates at 10ms2 as it is driven down the road. By the end of the route, the

Dmitry [639]

Answer:

$2.5\ m/s^2$

Step-by-step explanation:

Newton's Second Law: Force on a body is equal to the product of mass and acceleration of the centre of the mass of the body.

$Force(F)=Mass(m)\times Acceleration(a)$

Initially:

$Let\ mass=m\ kg\\\\acceleration=10\ m/s^2\\\\Force(F)=m\times 10=10m$

At the end of the road:

$Let\ mass=4\times m\ kg\\\\acceleration=a\ m/s^2\\\\Force(F)=4m\times a=4ma$

$Both\ forces\ are\ equal\\\\4ma=10m\\\\divide\ both\ the\ sides\ by\ m\\\\4a=10\\\\a=\frac{10}{4}\\\\a=2.5\ m/s^2$

8 0

2 years ago

Is y=8200(1.19)x a growth or decay and what is the percentage

Bad White [126]

Answer:

Growth 119

Step-by-step explanation:

It is positive and the percentage is 1.19 with the decimal moved 2 places to to the right.

6 0

2 years ago

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What is the amount of snow that can be placed in this globe if it is 4.7 inches wide?

zepelin [54]

Answer:

The awnswer is D. 54.3in^3

Step-by-step explanation:

7 0

2 years ago

If x=2 calculate the value of x cubed + x

Montano1993 [528]

X cubed (³) + x

x³ + x Plug in 2 for x
2³ + 2 Cube 2 (2 x 2 x 2)
8 + 2 Add
10

4 0

2 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

2 years ago

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