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Feliz [49]
3 years ago
8

A square has a side length of 3x3y5. What is the area of the square?​

Mathematics
1 answer:
kotykmax [81]3 years ago
8 0

Answer:

9x^3 y^5

Step-by-step explanation:

Since the formula to finding area is (L*W = A), we first have to find the width.

Width: 3x^3 y^5 times 2 = 6x^3 y^5

Then we add them. So, 3x^3 y^5 + 6x^3 y^5 = 9x^3 y^5

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Please help!<br> Subtract h + 3 ​​from​ 6h + 1.
cricket20 [7]
6h+1-h+3
=5h+4
because 6h-1h is 5h and 1+3 is 4
4 0
3 years ago
A friend has a 83% average before the final exam for a course. That score includes everything but the final, which counts for 25
Klio2033 [76]

Answer:

\mathrm{Best\:course\:grade\:possible:\:}87.25\%,\\\mathrm{Minimum\:score\:on\:final\:to\:earn\:at\:least\:a\:75\%\:for\:the\:course:\:}51\%

Step-by-step explanation:

Assuming the maximum score for the final is 100\%, we can multiply each score by its respective course weight and add them together to give a final score. If your friend did receive this maximum score of 100\%, their overall grade for the course would be:

83(1-0.25)+100(0.25)=\fbox{$87.25\%$}.

To find the minimum score they need to earn a 75% for the course, we set up the following equation:

83(1-0.25)+x(0.25)=75, where x is the minimum score she needs.

Solving, we get:

62.25+x(0.25)=75,\\x(0.25)=12.75,\\x=\fbox{$51\%$}.

8 0
2 years ago
Anybody help me please
GrogVix [38]

Answer:

Rotation 180° counterclockwise about the origin

7 0
3 years ago
Read 2 more answers
The seventh term of an arithmetic progression is equal to twice the fifth term. The sum
mario62 [17]

Answer:

a₁ = - 24

Step-by-step explanation:

The n th term of an AP is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₇ = 2a₅ , then

a₁ + 6d = 2(a₁ + 4d) = 2a₁ + 8d ( subtract 2a₁ + 8d from both sides )

- a₁ - 2d = 0 → (1)

The sum to n terms of an AP is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ]

Given S_{7} = 84 , then

\frac{7}{2} (2a₁ + 6d) = 84

3.5(2a₁ + 6d) = 84 ( divide both sides by 3.5 )

2a₁ + 6d = 24 → (2)

Thus we have 2 equations

- a₁ - 2d = 0 → (1)

2a₁ + 6d = 24 → (2)

Multiplying (1) by 3 and adding to (2) will eliminate d

- 3a₁ - 6d = 0 → (3)

Add (2) and (3) term by term to eliminate d

- a₁ = 24 ( multiply both sides by - 1 )

a₁ = - 24

6 0
3 years ago
Please solve for g!!!!
laiz [17]
The answer to the question

3 0
3 years ago
Read 2 more answers
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