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defon
2 years ago
12

Please I need help!!

Mathematics
1 answer:
nata0808 [166]2 years ago
7 0

The largest  possible complete design that can be painted on a 24’ x 16’ wall is 252' by 189'

<h3>Scale drawing</h3>

Scale drawing involves drawing a drawing in a different size and dimension that is proportional to the original dimension

<h3>Dimensions</h3>

The dimensions of the design for the farm scene painting is given as:

12" \times 9"

The dimensions of the wall is given as:

24' \times 16'

Convert feet to inches

24' \times 16' = (24 \times 12") \times (16 \times 12")

24' \times 16' = (288") \times (192")

Divide the dimensions of the wall by the painting to get the scale ratios (k)

k = \frac{288}{12}

k = 24

Also, we have:

k =\frac{192}{9}

k =21.33

Remove decimal (do not approximate)

k =21

By comparison;

21 is less than 24

Multiply 21 by the dimensions of the painting

So, we have:

Length = 12 \times 21 = 252

Width = 9 \times 21 = 189

Hence, the largest  possible complete design that can be painted on a 24’ x 16’ wall is 252' by 189'

Read more about scale drawings at:

brainly.com/question/810373

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Answer:

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Step-by-step explanation:

When there are 2 neatives together like that it changes to a positive. So it woud be 329+104=433

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multiply 14 x 2.54

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What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
3 years ago
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