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3 years ago

Calculate the speed for a car that went a distance of 125 miles in

Mathematics

1 answer:

Colt1911 [192]3 years ago

5 0

Answer:

To calculate spped you divide the distance by time so 125 divided by 2 will give you 62.5 m/s

Step-by-step explanation:

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How to get an equivalent answer to 1/4

jasenka [17]

The answer to this equation is 1/2

6 0

3 years ago

4. 30% to 50% of the money is for frozen food, a

strojnjashka [21]

Answer:

Step-by-step explanation:

It said the greatest amount,

So change 50% to .50

120x.50=60

8 0

3 years ago

Find the solution in slope-intercept form y+7=-3(x-1) and 3x+y=-4

Svetach [21]

Answer:

The slope intercept form of both given equations is : y = - 3 x - 4.

Step-by-step explanation:

Here, the given equations are:

y +7 = -3 ( x - 1 )

and 3 x + y = - 4

Now,the SLOPE INTERCEPT FORM of any given equation is given as:

y = m x + C : here, C = Y - intercept, m = Slope

Consider equation (1):

y +7 = -3 ( x - 1 ) ⇒ y + 8 = - 3 x + 3

or, y = -3x + 3 - 7 = -3x - 4

⇒ y = -3x -4

Hence, the slope-intercept form of the given equation is y = -3x -4.

Consider equation (2):

3 x + y = - 4 ⇒ y = -4 - 3 x

⇒ y = -3 x - 4

Hence, the slope-intercept form of the given equation is y = -3x -4.

6 0

3 years ago

A semicircle is cut out of a rectangular paperboard 24in long and 18n wide, as shown below. What is the perimeter of the paperbo

svet-max [94.6K]

Answer:

Perimeter of the paperboard that remains after the semicircle is removed = 94.26 in

Step-by-step explanation:

Watch the attached figure of how the semi circle is cut out of the rectangular paperboard.

Length = 24 in

Width = 18 in

Radius of the semi circle = Half of the width of the paperboard = $\frac{18}{2}$ = 9 in

1) Circumference of the semi circle = π*radius

= 3.14*9

= 28.26 in

2) Perimeter of the paperboard that remains after the semicircle is removed

= Top + Left + Bottom + Right Circumference of the semi circle

= 24 + 18 + 24 + 28.26

= 94.26 in

5 0

3 years ago

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^

guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

$d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j$

B )

Acceleration a is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j$

Hence , this is the required solution .

5 0

3 years ago