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andrew-mc [135]
3 years ago
12

The points (-9,f) and (-10,5) has a slope of 3. What is the value of f?

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0
Slope formula: (y2-y1)/(x2-x1)
(5-f)/(-10-(-9)) = 3
(5-f)/(-1) = 3
5 - f = -3
-f = -8, f = 8
Solution: f = 8
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Division is the opposite from multiplication
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3 years ago
PLZZZZZ HELLLLPPPP I NEEEEDDDDDD AAASSSSAAAPPPP<br> reward if u an help me ;)
Igoryamba

Answer:

1. x= -56.25

Expand

19.5-6.5x+36=201-4.5x-33

Simplify

-6.5x+55.5=201-4.5x-33

Simplify again

6.5x+55.5=-4.5x+168

Add 6.5 to both sides

55.5=-4.5x+168+6.5x

Simplify

55.5=2x+168

Subtract

55.5-168=2x

Simplify

112.5=2x

Divide both sides by 2

−112.5÷2 = x

Simplfy

x = -56.25

2. x > -7 ÷ 4

Or

Decimal Form: -1.75

Remove parentheses

12x>4x+5−19

Simplify

12x>4x-14

Subtract

12x-4x>-14

Simplify

8x>-14

Divide both sides by 8

x > -14 ÷ 8

Simplify

x > -7 ÷ 4

Or

Decimal Form: -1.75

3. Answer: Step 2 has an error

Step-by-step explanation:

Given equation,

2(10 - 13x) = -34x + 60

By distributive property,

20 - 26x = -34x + 60

Now, we need to isolate x on the left side of the equation,

For this we need to eliminate constant term from the left side,

20 will be eliminated by subtracting 20 from both sides ( subtraction property of equality )

I.e. Step 2 has an error,

We need to use subtraction property of equality instead of using addition property of Equality,

Note : The correct steps would be,

Step 2 : 20 - 26x = -34x + 60 ( Subtraction property of equality )

Step 3 : 8x = 40  ( addition property of Equality )

Step 4 :  x = 5  (  Division Property of Equality )

Hope this helps!!! Good luck!!! ;)

8 0
3 years ago
determine the slope-intercept form of the equation of the line parallel to y=-4/3x + 11 that passes through the point (-6,2
Zielflug [23.3K]

is it...

y=-4/3x+10

........................

3 0
2 years ago
Desiree ran 12 miles in 1 hr, 48 minutes, What was Desiree's average running rate, in minutes per mile
Rina8888 [55]

Answer: 1/9

Step-by-step explanation:

60 + 48 = 108 minutes

12 / 108 = 1/9

3 0
3 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

5 0
3 years ago
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