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3 years ago

If you are driving 60km/h and you speed up to 70km/h over a period of 5 seconds, what is

Mathematics

2 answers:

lbvjy [14]3 years ago

8 0

Answer:

Step-by-step explanation:

solution:

Data,

driving speed of car intially velocity at = Vi = 60km/h

the speed was then raised to final velocity at = Vf = 70km/h

the total period of speed raised = t = 5 sec

Req, acceleration = ?

put the formulae : a = Vf - Vi/t

a = 70 - 60/5

a =<em> 2 m/sec2 ans</em>

/ means divided by,

quester [9]3 years ago

5 0

Answer:

fast as heck 0.o (i think 170

Step-by-step explanation:

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-8x + 8y =24<br> 8x +5y =28

Hunter-Best [27]

Answer: (1,4)
Hope this helps! Good luck!

7 0

3 years ago

A rectangular swimming pool is surrounded by a deck as shown. The area (in square feet) of the pool is represented by x^2-10x+16

maw [93]

Answer:

Step-by-step explanation:

First we need to factorize the given area of the rectangle as shown;

Given

A(x) = x^2-10x+16

Factorize

A(x) = x²-8x-2x+16

A(x) = (x²-8x)-(2x+16)

A(x) = x(x-8)-2(x-8)

A(x) = (x-2)(x-8)

Since area of a rectangle = length ×width

Hence the binomial that represents the width of the pool is w(x) = x-2

b) If the width of the deck is 17feet

Width = x-2

17 = x-2

x = 17+2

x = 19feet

Get the length

Length = x-8

Length = 19-8

Length = 11

Get the perimeter

P = 2(L+W)

P = 2(17+11)

P =2(28)

P = 56feet

Perimeter of the deck is 56feet

8 0

3 years ago

1. Nikola invested $6.000 in a 3 year

Kamila [148]

I would rather say A or C

3 0

2 years ago

Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due

ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

Formula : d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

Formula : D= $\sqrt{(x2-x1)^2+(y2-y1)^2}$

⇒ $D = \sqrt{20^2t^2+15(t-1)^2}$

Now let $F(t) = D^2(t)$

∵ $F'(t) = 800t + 450(t-1) = 1250t -450\\F'(t) =0$

⇒ t= 450/1250

⇒ t= .36 hours

⇒ = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

3 0

3 years ago

Susan solved 200-91 and decided to add her answer to 91 to check her work explain why this strategy works

zhuklara [117]

This strategy works because if you subtract one number from another the sum of the two numbers should be the that is being subtracted from. (In this case, it's 200)

3 0

3 years ago

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