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Vikki [24]
3 years ago
11

Chloe is painting on a stretched canvas that has an area of 108 square inches. The length of the canvas is 12 inches. The width

of the canvas is how many inches. Chloe paints a 1-inch-wide blue border on the canvas. The area left for painting after painting the border is how many square inches. I WILL GIVE BRAINLIEST!
:-)

Mathematics
1 answer:
Stolb23 [73]3 years ago
6 0

Answer: Width of canvas = 9 inches

Area left for painting after painting the border = 70 square inches.

Explanation :

Since we have given that

Area of painting = 108 square inches

Length of canvas = 12 inches

As we know that

\text{Area of rectangle} = length\times breadth

108=12\times breadth\\\\\frac{108}{12}=breadth\\\\9=breadth

So, breadth of canvas = 9 inches .

Now, Chloe paints a 1 inch wide blue border on the canvas.

\text{So, remaining length left for painting after painting the border} = 9-1-1=9-2=7\text{ inches}

\text{ Remaining breadth left for painting after painting the border }= 12-1-1=12-2= 10\text{ inches}

\text{So , area of remaining portion }= length\times breadth \\\\ \text{ area of remaining portion }=7\times 10\\\\\text{ area of remaining portion }=70 \text{ square inches}

Hence, remaining area = 70 square inches .

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mr Goodwill [35]

Answer and Step-by-step explanation:

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6 0
3 years ago
HELP PLEASE IM REALLY STUCK
liberstina [14]
The 15th term will be 71. Why? Well, see below for an explanation!

By subtracting all of these numbers by the term that comes prior to them, we will find that all of them result in 5. Because of this, we know that each time the term increases, 5 is being added to the numbers. Additionally, I noticed that all of the numbers in this arithmetic sequence only end in a 1 or a 6. Because of this, we can apply the same principle when adding 5 each time:

First term: 1
Second term: 6
Third term: 11
Fourth term: 16
Fifth term: 21
Sixth term: 26
Seventh term: 31
Eighth term: 36
Ninth term: 41
Tenth term: 46
Eleventh term: 51
Twelfth term: 56
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Fourteenth term: 66
Fifteenth term: 71

By adding 5 each time and keeping in mind that the digits all end in only 1 or 6, we will find that the fifteenth term results in 71. Therefore, the 15th term is 71.

Your final answer: The 15th term of this arithmetic sequence comes down to be 71. If you need extra help, let me know and I will gladly assist you.
4 0
2 years ago
What percent of the values are between 10 and 30? Round to the nearest tenth of a percent if necessary.
skad [1K]

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6 0
3 years ago
Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
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ANSWER

See below

EXPLANATION

Part a)

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b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

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(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

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f''(0) \: =\: 0

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3 years ago
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OLga [1]

Answer:

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Let  P( B )  be the  probability that a customer pays late each month

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The probability that a customer pays on-time and late each month is  P(A n B) and  the value is  zero ( 0 ) given that it is impossible

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=>     0.82  =  0.55 + P( B ) -   0

=>    P(B)  =  0.27

8 0
3 years ago
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