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omeli [17]
2 years ago
11

Please help will give brainliest

Mathematics
1 answer:
ivanzaharov [21]2 years ago
3 0

Answer: 1 and 7/15

Step-by-step explanation:

I made the common donominator 15 and then did the problem but the donominator stays the same

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Can you please help me
miskamm [114]

Answer:it might be D

Step-by-step explanation:

3 0
3 years ago
Find the slope of line passing through the points (3,2) and (2,3) Use slope formulation show your work
mart [117]

Answer:

the slope is -1

Step-by-step explanation:

the slope formula is y^2 - y^1 divided by x^2 - x^1

y^2 is 3

y^1 is 2

x^2 is 2

x^1 is 3

6 0
3 years ago
HELP ASAP PLEASE!
Genrish500 [490]

Step-by-step explanation:

lets take a look....

In order to find the perimeter of a triangle you have to get the sum of all sides. In other words,

Blue triangle:

p = (4x +2) + (7x + 7) + (5x  - 4) \\ p = 4x + 2 + 7x + 7 + 5x - 4 \\ p = 16x + 5

Red triangle:

p = (x + 3) + (2x - 5) + (x + 7) \\ p = x + 3 + 2x - 5 + x + 7 \\ p = 4x + 5

Difference:

d = (16x + 5) - (4x + 5) \\ d = 16x + 5 - 4x - 5 = 12x \\ d = 12x

4 0
3 years ago
Find the surface area of the cube shown below. units 2 2 squared
ruslelena [56]

The surface area of the cube with a side length of 2 units is 24 units squared

<h3>How to determine the surface area?</h3>

The side length of the cube is given as:

l = 2

The surface area is calculated as:

Surface area = 6l^2

This gives

Surface area = 6 *2^2

Evaluate

Surface area = 24

Hence, the surface area of the cube is 24 unit squared

Read more about surface area at:

brainly.com/question/13175744

#SPJ1

7 0
2 years ago
Could you have a polynomial with solutions 2i and 5i? If not explain why not. If yes, then construct the polynomial.
baherus [9]

If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi.  Imaginary solutions always come in pairs.

So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.

(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.

(x-2i)(x+2i)(x-5i)(x+5i)  = (x^2+4)(x^2+25)

                                   = x^4 + 29x^2 + 100

So the equation would be x^4 + 29x^2 + 100 = 0.

Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.

5 0
3 years ago
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