Velma Said:

math geometry problems best answer 10 pts?

We Answered:

1)

16² + 30² = hypotenuse²

256 + 900 = hypotenuse²

1156 = hypotenuse²

34 = hypotenuse

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2)

15² = 12² + x²

15² - 12² = x²

225 - 144 = x²

81 = x²

9 = x = other side

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3)

diagonal of a square = side * sqr (2) = 4 * sqr (2)


explanations :

diagonal² = side² + side² ===> the diagonal is the hypotenuse of a right triangle

diagonal² = 2 side²

diagonal = (side) sqr (2)

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Russell Said:

PLEASE ANSWER high school level Geometry math problems!!!?

We Answered:

1. AAS because

the right angle
the vertical angle in the middle
and then the midpoint creating 2 congruent lines

2. SAS because

PR is congruent to SQ
the right angles
both share RQ

3. AASbecause

EC would be congruent to AF
vertical angles
<ECG woud be congruent to <FAG

4. m <AEC would be 80 degrees

becuase <AEC is congruent to <BED since they are vertical angles and you would put them equal to each other

4x - 40 = x + 50
3x - 40 = 50
3x = 90
x = 30

then you plug in 30 in the equation for <AEC

<AEC = 4 (30) - 40
120 - 40
<AEC = 80 degrees

HOPE THIS HELPS!!!!

Sherry Said:

Challenging geometry math problems! Help, please!?

We Answered:

1. x(1+sqrt(2) ) = 12

x = 4.97 inch


2. AB = arctan(4/3) = 53.1 degree

Jeremy Said:

Can someone help me with this math problem - Algebra 1: Geometry Problems?

We Answered:

First, you need to define some of your variables.

You know that A (Area) is equal to l x w (length times width).
So, A = l x w.

You know that A = 33m^2. You don't need to worry about the m^2 right now, so leave A = 33.

Now, the length is 5m greater than twice the width. To write this algebraically, you would write:
l = 2w + 5

Substitute the value of l in and you get:
(2w + 5)*(w) = 33

Multiplied out, that's 2w^2 + 5w = 33.
Subtract 33 and you get 2w^2 + 5w - 33 = 0.

Factor that and you get (2w + 11)*(w - 3).

Since w cannot be negative, your only zero left is w = 3.

Since l = 2w + 5, substitute 3 in and you get l = 6 + 5, which equals 11.

11x3 = 33.

Happy to help.

Brandon Said:

Geometry MATH PROBLEMS?

We Answered:

In rectangular form (components) you would write the vector:
(11 cos(30), 11 sin(30))
and then substitute the values for the sin and cos.

The exact value (rather than a rounded one) is:
(11 sqrt(3) / 2, 11 / 2).

Alternatively, using i / j notation, you could write:
[ 11 sqrt(3) / 2 ] i + [ 11 / 2 ] j.

By adding the squares of the components and taking the square root, all you have done is get back to the original magnitude (11) with a rounding error. You don't need that in the answer. The components alone are sufficient.

The lateral surface area of a cone is:
pi r s
where r is the radius of the base and s is the slant height.

As the circumference is 40m, the radius is (40 / 2pi) metres, giving lateral surface area:
pi (40 / 2pi) 38
= 20 * 38
= 760 m^2.

The area cannot be expressed in metres -- it has to be square units.

Craig Said:

2 Geometry Math Homework Problems?

We Answered:

1. That's Gauss' formula: you know that 1+100=101, 2+99=101, 3+98=101, etc. So you have that (100+1)(100/2).
2. 1, 3, 6, 10, 15, 21