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3 years ago

5

The temperature at 9 A.M. was 42°F. The next four hours the temperature dropped 2 degrees per hour. What was the temperature at

1 P.M.?

2 answers:

Zinaida [17]3 years ago

7 0

Answer:

34

Step-by-step explanation:

2x4 = 8, 42-8 = 34

Varvara68 [4.7K]3 years ago

5 0

Answer:

34 degrees farenheit yeee

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A toy train has 8 compartments and an engine that is 8 inches long. Each compartment is 4 inches long. Each compartment is joine

luda_lava [24]

Answer:

64in

Step-by-step explanation:

links=8*3=24

compartments + engine=4*8+8=40

40+24=64

3 0

3 years ago

Katie said,"Negative numbers are integers."What was her error?

rodikova [14]

Integer means a whole number , a number that is not fraction
Katie did not specifically said what numbers , the numbers can be fraction and it would not be called integer so that is Katie's error

4 0

3 years ago

Answer while showing work I need u to show ur work and answer

skad [1K]

Answer:

7cm

Step-by-step explanation:

One side of a square is increased by 8 cm, let us say the length got increased by 8 cm. So we have length of x + 8 cm

An adjacent side, the width, decreased by 2 cm. So we have width of x - 2cm

Perimeter of rectangle = 2(length + width)

40 = 2( (x + 8) + ( x - 2) )

40 = 2( 2x + 6)

40 = 4x + 12

28 = 4x

x = 7 cm

That is 7 cm of the length of the side of the square

4 0

3 years ago

What is the difference, in hours and minutes, between these times?

solmaris [256]

Answer:

a)3hours and 15mins

b)3hours and 10mins

c)2hours and 20mins

d)8hours and 40mins

3 0

3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago