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Zanzabum
3 years ago
10

What is the area of the figure?

Mathematics
2 answers:
mash [69]3 years ago
5 0

Answer:

<h2>64.26 cm².</h2>

Explanation is given in the above photo.

egoroff_w [7]3 years ago
4 0

Answer:

64.26

Step-by-step explanation:

6 times 6 is 36 and 28.26 is the area of of the 1/4 circle

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Factorize (x-y)^2+(x-y)​
olga2289 [7]

..............answer..

5 0
2 years ago
Compare the decimals 16.30 and 16.3
pshichka [43]
16.30 and 16.3 equal. If you ever have a problem like that then you just add on a zero.

Example:
19.4500 = 19.45 Just add two zeros on the end
19.45 + two zeros = 1.4500
8 0
3 years ago
What is the width of it
notka56 [123]
The width of the rectangle is 16cm. this is because 16x4 = 64. so the height of the rectangle is 64 cm and 16cmx64cm = 1024cm squared. which is why the width of the rectangle is 16cm.
4 0
2 years ago
15. Apply Mathematics (1)(A) Suppose you have a box with a 4 X 4-in.
erastova [34]

Answer:

Step-by-step explanation:

Formula for calculating the surface area of the box S = 2(LW+LH+WH) where

L is the length of the box

W is the width of the box

H is the height of the box

If the box is square based with dimension 4 * 4in, then L = W = 4in. substituting this values given into the formula we will have;

S = 2(4(4)+4H + 4H)

S = 2(16+8H)

S = 32+16H

<em>Hence, The function that represents the surface area of this box as a </em>

<em>function of its height is S = 32+16H where H is the height of the box</em>

<em></em>

Given H = 6.5in, to evaluate the function, we will substitute h = 6.5in into the modeled equation;

S = 32+16H

S = 32+16(6.5)

S = 32+106

<em>S = 138in²</em>

<em>Hence the total surface area of the box is 138in²</em>

5 0
2 years ago
1.) Determine the type of solutions for the function (Picture 1)
NNADVOKAT [17]

Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

∴ 6 = b - 2·b + 8 = -b + 8

6 - 8 = -b

∴ -b = -2

b = 2

b = 4·a

∴ a = b/4 = 2/4 = 1/2

The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

2 nonreal complex roots

2) Give that the graph of the function touches the x-axis once, we have;

1 Real Solution

3) The given function is f(x) = 2·x² + 8·x + 6

The general form of the quadratic function is f(x) = a·x² + b·x + c

Comparing, we have;

a = 2, b = 8, c = 6

The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

4.) The given function is g(x) = (-2/5)·(x - 4)² + 6

The parent function of a quadratic equation is y = x²

A vertical translation is given by the following equation;

y = f(x) + b

A horizontal to the right by 'a' translation is given by an equation of the form; y = f(x - a)

A vertical reflection is given by an equation of the form; y = -f(x) = -x²

A narrowing is given by an equation of the form; y = b·f(x), where b < 1

Therefore, the transformations of g(x) from the parent function are;

g(x) is a reflection of the parent function, with the graph of g(x) being narrower by 2/5 than the graph of the parent function. The graph of g(x) is shifted right by 4 units and is then slides up by 6 units.

7 0
2 years ago
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