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3 years ago

Which polygon has perpendicular sides?

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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Mr. Jones charged Roberta $32 for parts and $15 per hour for labor to repair her

Galina-37 [17]

Answer:

Step-by-step explanation:

Since 3 hours and $15 an hour

5 0

3 years ago

What is the sum in simplest form?<br> 41 +13 = 1

Alborosie

Answer:

x + y = 41

x - y = 13

------------------

2x = 54..............We divide 2 such that

x = 27

Then we plug 27 into the equation such that

27 + y = 54.................Subtract 27 to get

y = 14

Step-by-step explanation:

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3 0

3 years ago

Evaluate 5-\dfrac{t}35−

solmaris [256]

Answer:

31.438

Step-by-step explanation:

yan ppo ang tama sagot

7 0

3 years ago

Read 2 more answers

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

2 years ago

U + 3b - 2a + 2 solve for a

MA_775_DIABLO [31]

Answer:

a=2

Step-by-step explanation:

u=3b-2(2) or u=3b-4

3 0

3 years ago