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matrenka [14]
3 years ago
8

Jeremy purchases a rectangular canvas to paint a portrait. The canvas has a length of (4x + 5) inches and a width of (x + 10) in

ches. He paints a black border around the edge of the canvas, which has an area of (x2 + 5x + 4) square inches. How much area is available for Jeremy to paint the portrait?
Mathematics
2 answers:
ozzi3 years ago
8 0

Answer:

(3x^2 + 40x + 46) square inches

Step-by-step explanation:

see answer

mariarad [96]3 years ago
7 0

Answer:

Step-by-step explanation:

25 ml hope I helepd thanks

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Consider this expression
Sloan [31]

Answer:

\displaystyle \large{(x + 9)(x - 8)}

The value of a is 9 and b is -8

Step-by-step explanation:

\displaystyle  \large{(x + a)(x  + b) =  {x}^{2}  + (a + b)x + ab}

We need to find two numbers that add up or subtract off to 1.

Then we also use the same two numbers that multiply and get -72.

Let's see:-

  • 9 and -8 seem like correct values.

Because 9+(-8) is 9-8 = 1

And 9(-8) is -72.

Therefore:

\displaystyle \large{ {x}^{2}  + x - 72 = (x + 9)(x - 8)}

The value of a is 9 and the value of b is -8 according to form of (x+a)(x+b)

8 0
3 years ago
Read 2 more answers
How would you get answer for at the grocery store cherries cost $5 for 2 pounds and $7.50 for 3 pounds. What is the cost per pou
goldfiish [28.3K]
Hope this answer helps you! If it did, please mark it as the brainiest! I would really appreciate it! :)

8 0
3 years ago
a trapezoid has parallel sides of length 26 mm and 44 mm. its other sides measure 10 mm and 17 mm. what is the length of the bim
kirill115 [55]

Answer:

35 mm.

Step-by-step explanation:

That would be (26 + 44)/2

= 70/2

= 35 mm.

5 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
NEED ANSWER ASAP BRIANLIEST TO GIVE A house increases in price by 3% each year. If the house starts at £180,000, how much is it
kow [346]

Answer:

£196,690.86 or £196,691

4 0
4 years ago
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