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3 years ago

Would anyone please be able to help me??

Mathematics

2 answers:

Gekata [30.6K]3 years ago

8 0

Answer:

a. = 111 inches squared b.= 506 inches squared c.=143 inches squared

Step-by-step explanation:

To find the area I broke the shape into two parts one where you multiply 7 and 9 to get the area and one where you multiply 12 and 4 to get the area. The first area is 63 and the second is 48 and together the entire area is 111 inches squared

To find the area here you do the same thing as a. you break the shape into two parts a part where you multiply 29 and 14 and a part where you multiply 20 and 5. In all you get 506 inches squared.

I this problem you find the area of the triangle and the rectangle.The formula of the triangle is b times h times 1/2 in this case. 6 times 13 times 1/2 which is 39. the area of the rectangle is 8 times 13 = 104. So the total area is 143

Sorry i took so long I had to do something

Svetllana [295]3 years ago

4 0

Answer:

Area of first shape = 97cm^2

Area of second shape = 506cm^2

Area of third shape =143 cm^2

First shape :

i. Area if rectangle in the fig. = l*b

= 12*4

= 48 cm^2

Now, in the same fig. we construct a line and make it a square,

ii. Area of square = s*s

= 7*7

= 49 cm^2

Area of whole shape = 48cm^2 + 49cm^2

= 97 cm^2

Second Shape :

i. Area of the rectangular shape(above one)

= l*b

= 29*14

= 406 cm^2

ii. Area of the rectangle (stick like shape)

= l*b

= 20*(29-24)

= 20*5

= 100 cm^2

Area of whole shape = (100 + 406) cm^2

= 506 cm^2

Third shape :

i. Area of the triangle = 1/2 *b* h

= 1/2 * 13 * 6

= 39 cm^2

ii. Area of the square = s*s

= 13*8

= 104 cm^2

Area of whole shape = (104 + 39) cm^2

= 143 cm^2

Hope it helps!

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Leno4ka [110]

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3 0

4 years ago

Identify Relationships. Do the numbers show a pattern? 11.23,10.75,10.3,9.82,9.37,8.89

Alexxandr [17]

<h3>Answer:</h3>

Yes. The pattern is predicted by the formula ...

f(x) = -0.465x +11.68 +0.015·mod(x, 2)

<h3>Step-by-step explanation:</h3>

Initial investigation

The predictions from a line of best fit alternate being too low and too high. That is to say, every other point fits perfectly on a line, and the alternate points fit perfectly on a parallel line.

Line through alternate points

If we use x=1, 2, 3, ... for the number of the member of the sequence, then we have ....

(x, f(x)) = (2, 10.75), (4, 9.82), (6, 8.89)

These points describe a line that can be found using the two-point form of the equation for a line:

y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

y = (9.82 -10.75)/(4 -2)·(x -2) +10.75 . . . . . using the values from the first two even-numbered points

y = (-0.93/2)·(x -2) +10.75 . . . . . . . . . . simplify a bit

y = -0.465x +11.68 . . . . . . . . . . . . . . . . simplify to slope-intercept form

Correction for other points

We know that this predicts a value too low when x is odd. We can find the amout the prediction is low and add a conditional term to make up the difference.

For x = 1,

y = -0.465·1 +11.68 = 11.215

The actual value is 11.23, or 11.23 -11.215 = 0.015 higher than the equation predicts. We can use the modulo function to give a value that is 1 for odd numbers and 0 for even numbers. For odd values of x, we can add this difference to make the formula predict every number in the sequence.

f(x) = -0.465x +11.68 +0.015·mod(x, 2) . . . . equation for the given points

_____

The table attached shows the given points (y1) and the values predicted using the above formula (f(x1)). (The x1 and y1 used in the attachment are the names of the lists of x-values and y-values. They are not related to the x1 and y1 used above in the 2-point form of the equation for a line.)