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tensa zangetsu [6.8K]
3 years ago
8

So I need help with this math problem​

Mathematics
1 answer:
jenyasd209 [6]3 years ago
4 0
You would solve this problem by using Pythagorean Theorem. This theorem states a^2+b^2=c^2. C is ALWAYS the hypotenuse. We would have to solve for the smaller triangle first. It would look something like this.
3^2+b^2=5^2
9+b^2=25
b^2=16
b=4
Therefore, the missing side to the small triangle is 4. Now, we can solve for the hypotenuse of the bigger triangle,
4^2+7^2=c^2
16+49=c^2
65=c^2
Side z is approximately equal to 8.
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3 years ago
Which statements about angles are true? Check all that apply.
Vadim26 [7]

Answer:

A,C,D,E

Step-by-step explanation:

The first one is true because in fact, supplementary angles' measures add up to 180°.

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acute angle-an angle between 0 and 90 degrees.

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Descried how to derive the quadratic formula from a quadratic equation in standard form
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Answer:

The standard form of a quadratic equation is:

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Quadratic Formula Derivation:

ax^2+bx+c=0\\

$x^2+\frac{b}{a}x+\frac{c}{a} =0 $

$x^2+\frac{b}{a}x = -\frac{c}{a}$

Completing the Square:

$x^2+\frac{b}{a}x +\frac{b^2}{4a^2} = \frac{b^2}{4a^2}-\frac{c}{a}$

$  ( x+\frac{b}{2a} )^2 =  \frac{b^2-4ac}{4a^2}  $

Square Root property:

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

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

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

8 0
3 years ago
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bonufazy [111]
The basket ball players because the mean is higher!! Good Luck!
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4 years ago
Read 2 more answers
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