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DanielleElmas [232]
3 years ago
9

Which graph represents the solution for the equation 1/6x=-2x+5

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
8 0
More information please.
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An auto weighing 2,500 pounds is on a street inclined at 10 with the horizontal. Find the force necessary to prevent the car fro
Misha Larkins [42]

Answer:

1931.7N

Step-by-step explanation:

We are told in the question that :

An auto ( a car) weighs = 2500 pounds

It s inclined at n horizontal Ange of 10°

We are asked to find the force that would prevents it from rolling down the street.

Since the unit for Force = Newton or kgm/s²

Step 1

Convert Weight in pounds to kg

1 pound = 0.453592kg

2500 pounds =

2500 pounds × 0.453592kg

= 1133.981kg

Step 2

Find the force that would prevents it from rolling down the street.

Force = Mass × Acceleration due to gravity × sin θ

Acceleration due to gravity = 9.81m/s

Force = 1133.981kg × 9.81 × sin 10°

Force = 1931.7237321 N

Approximately = 1931.7N

4 0
3 years ago
1.Graph f(x) = -1.5x +6<br> 2.Graph f(x) = -1/2x-5
nalin [4]

Ans(1):

Given equation is f(x)=-1.5x+6

we can plug any number like x=0 and x=2 to find the f(x) also called y-value

plug x=0

f(x)=-1.5x+6 =-1.5*0+6 =0+6 =6

Hence first point is (0,6)

plug x=2

f(x)=-1.5x+6 =-1.5*2+6 =-3+6 =3

Hence first point is (2,3)

now we can graph both points then join them to get final graph of f(x)=-1.5x+6

---------------------

Ans(2):

We can repeat exactly same process for f(x) = -1/2x-5.

So the final graph will look like attached picture:


5 0
2 years ago
3.1 Find the equation of the line passing through (4, 5) and parallel to the line 3x – 2y = 4.
valentinak56 [21]

                   \rule{50}{1}\large\blue\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}

            <em>Find the equation of the line passing through (4, 5) and parallel to </em>

<em>                 3x-2y=4.</em>

<em />

<em>  </em>\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

First of all, We need to convert this equation from standard form (ax+by=c) into slope-intercept form (y=mx+b).

\rightarrow First, add 2y on both sides:-

                            \Large\textsf{3x+2y=4}

\rightarrow Now, subtract 3x on both sides:-

                  \Large\textsf{2y=-3x+4}

\rightarrow Divide by 2 on both sides

    \Large\text{$y=-\displaystyle\frac{3}{2}x +2$}

\rule{300}{3}

Now, let's find the equation of the line that's parallel to the given line.

Remember, if lines are parallel to each other, then their slope's the same.

So the slope of the line that's parallel to the line whose equation we just found, is

\Large\text{$-\displaystyle\frac{3}{2}$}

Now let's write the equation in point-slope form:-

\hookrightarrow\sf{y-y_1=m(x-x_1)}

Replace numbers with letters, as follows:-

\sf{y-5=-\displaystyle\frac{3}{2} (x-4)}

The equation above is the equation in point-slope form.

Incase you need it converted to slope intercept form, refer to the steps below ~

Multiply -3/2 times x and -4:-

\hookrightarrow\sf{y-5=-\displaystyle\frac{3}{2} x-6}

Now, Add 5 on both sides:-

\hookrightarrow\sf{y=\displaystyle\frac{3}{2} x+1}

\Uparrow\sf{Our\:equation\:in\:slope\;intercept\;form}

<h3>Good luck with your studies.</h3>

#TogetherWeGoFar

\rule{300}{1}

4 0
2 years ago
Sometime outiers can have a significant effect on statistics and sometime they don't . Use the following distribution to determi
IrinaVladis [17]
What is the test name?
8 0
3 years ago
Find the first four terms of the sequence given by the following.<br> an=51 + (n-1).8, n=1, 2, 3...
pav-90 [236]

Answer:

Step-by-step explanation:

a1 = 51 + (1 -1 ) * 0.8

a1 = 51

a2 = 51 + (2 - 1)*0.8

a2 = 51 + 0.8

a2 = 51.8

a3 = 51 + (3 - 1)*0.8

a3 = 51 + 2*0.8

a3 = 51 + 1.6

a3 = 52.6

a4 = 51 + (4 - 1)*0.8

a4 = 51 + 2.4

a4 = 53.4

5 0
3 years ago
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