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Rom4ik [11]
4 years ago
10

a car travels at an initial velocity of 6.0m/s and decelerates at -3.0m/s^2 how much time elapses until the car’s displacement i

s 5.3 m while traveling in the original direction
Mathematics
1 answer:
kotykmax [81]4 years ago
7 0

Answer:

b i think

Step-by-step explanation:

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Given the piecewise function shown below, select all of the statments that are true.
timurjin [86]
The <span>given the piecewise function is :
</span>
f(x) = \[ \begin{cases} &#10;      2x & x \ \textless \  1 \\&#10;      5 & x=1 \\&#10;      x^2 & x\ \textgreater \ 1 &#10;   \end{cases}&#10;\]

To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25


To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4

To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4

To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:<span>
\framebox {B. f(2) = 4} \ \framebox {D. f(1) = 5}</span><span></span>
8 0
4 years ago
Help me!! Thank you for the help!!
VLD [36.1K]

Answer:

D. 0.34

Step-by-step explanation:

0.24²+0.31²=x² then you find the square root

6 0
3 years ago
Read 2 more answers
If F(X)= 6X-4 and F(A)= 26 what is the value of a? Explain
sammy [17]
Easy if you know the formula

all this means is that your x value is a, and when it is, will give you the value 26.

f(a) = 6a - 4 = 26

6a = 30
a = 5
3 0
3 years ago
Read 2 more answers
Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).
alexira [117]

Let f(x,y,z)=z-\ln(x+y). The gradient of f at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

\nabla f(1,0,0)\cdot(x-1,y,z)=0

Compute the gradient:

\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)

Evaluate the gradient at the given point:

\nabla f(1,0,0)=(-1,-1,1)

Then the equation of the tangent plane is

(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}

7 0
4 years ago
Find the center and radius of this circle x^2+y^2-6x-10y+30=0
Vsevolod [243]

Answer:

The circle's centre is at the position (3, 5), and it has a radius of 2

Step-by-step explanation:

First let's put it in a useful format by completing the squares:

x² + y² - 6x - 10y + 30 = 0

x² - 6x + y² - 10y = -30

x² - 6x + 9 + y² - 10y + 25 = -30 + 9 + 25

(x - 3)² + (y - 5)² = 4

This tells us that the centre position is (3, 5) and the radius is √4, or 2

5 0
3 years ago
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