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Wewaii [24]
3 years ago
7

Does the following picture represent a linear function?

Mathematics
2 answers:
asambeis [7]3 years ago
8 0
Nope, it starts down, comes up a little and goes back down which makes it a parabola, not a line.
nekit [7.7K]3 years ago
7 0

Answer: No

Step-by-step explanation: For x 0-2, the y is increasing by +5 every time. However, the moment it hits x=2, it starts decreasing by -5 every time. This means that it does not follow on set pattern (either +5 or -5) and looks more like a ^ than a /.

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A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
3 years ago
How do you figure out the answer?
svp [43]

we can divide the composite shape into on quadrilateral and triangle as shown in the diagram. since the line that joins the two shape is parallel to the base AB, the angle DEC is 85°.The angle CEA is 180°- angle DEC = 180-85=95. Angle BCE is sum of angle CEA, ABC and EAB subtracted from360°= 360°-(95+95+85)=85°. Angle DCE is 120°- ECB=35°. So, angle CDE = 180°-(35°+85°)= 60°. For reference see the diagram.

7 0
3 years ago
This is due today, please help!!! I will give brainliest.
WINSTONCH [101]

Answer:

Step 1, all the exponents are increased by 4

Step-by-step explanation:

The first incorrect step occurred in Step 1, where all the exponents were increased by 4.

This is mathematically incorrect due to exponential rules. When distributing exponents inside parentheses, we have to multiply the existing exponents inside the parentheses by the exponent outside the parentheses.

For example, (x²)³ is not x²⁺³, but rather, x⁽²⁾⁽³⁾.

We multiply the exponents instead of adding them together.

Therefore, the correct Step 1 should multiply all the variables' exponents by 4.

Steps 2 and 3 are correct since we do add the exponents when multiplying exponents with the same base, and we do subtract exponents with the same base when dividing.  

6 0
3 years ago
I(1, 1) and J (-3,-3)<br> Find the midpoint of the line segment
Pavlova-9 [17]
The Answer is (-1,-1)
4 0
3 years ago
In the diagram below, value of x is: <br><br> A. 11 <br><br> B. 55<br><br> C. 5<br><br> D. 35
koban [17]
Sum of complementary angles = 90
so
5x + 35 = 90
5x = 90 -35
5x = 55
x = 11

answer is A. 
x = 11
3 0
3 years ago
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