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3 years ago

In 1989 a postage stamp cost $.25. In 1999 a postage stamp cost $.33. Find the average rate of

Mathematics

2 answers:

zalisa [80]3 years ago

7 0

Answer:

0.33-0.25= 0.08

then 1999-1989= 10

lastly 0.008

Step-by-step explanation: so the average rate of

change in postage in cents per year. is 0.8 of a cent. or 0.008 of a dollar.

Mrac [35]3 years ago

6 0

I’m pretty sure it is 80 cents, because it changed by 8 cents in 10 years so you’re supposed to divide 8 by 10 and get .8 which is equivalent to 80 cents

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A runner is traveling at a constant rate of 8 meters per second. How long does it take for the runner to travel 100 meters?

MrRa [10]

Answer:

Total time taken = 12.5 second

Step-by-step explanation:

Given:

Speed of runner = 8 m/s

Total distance cover = 100 m

Find:

Total time taken

Computation:

Time taken = Distance / Speed

Total time taken = 100 / 8

Total time taken = 12.5 second

3 0

3 years ago

Look at the Picture. Look at the Picture.

lys-0071 [83]

Answer:

325 square inches

Step-by-step explanation:

Consider the attachment below for further reference. Ideally we would split this figure into parts, and solve as demonstrated by the attachment. I have labeled each rectangle as rectangle 1, rectangle 2, rectangle 3 etc. ;

$Rectangle 1 Area = 17 in * 5 in = 85 square in\\Rectangle 2 Area = ( 17 in - 5 in ) * 14 in = 168 square in,\\Rectangle 3 Area = ( 12 in - 3 in ) * 8 in = 72 square in\\\\Total Area = 85 + 168 + 72 = 325 square inches$

Hope that helps!

5 0

3 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$