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1 year ago

I will give lots of points please help

Mathematics

2 answers:

r-ruslan [8.4K]1 year ago

6 0

Use Pythagorean theorem to confirm the right triangle:

AB² + BC² = AC²

12² + 5² = 13²
144 + 25 = 169
169 = 169

Confirmed

The triangles are similar:

ΔABC ~ ΔRST

Corresponding sides have same ratio:

AB/RS = AC/RT

Substitute known values and find the missing length:

12/24 = 13/RT
1/2 = 13/RT
RT = 13*2
RT = 26

creativ13 [48]1 year ago

3 0

Answer:

yes it is a right triangle if you look closely at triangle ABC there is a right triangle mark the little square

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3200 plus 1800 is 5000.
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So 36%

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3 years ago

P(x)=(x-2)(x-3)(x+7) find the constant term

hram777 [196]

Short method: multiply constants
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Long mothod: expand it
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3 years ago

T/0.3 = 1.7/0.9 ( the one I'm pointing at)

Vikentia [17]

I hope this helps you

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3 years ago

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Irina-Kira [14]

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2 years ago

Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real

hoa [83]

Answer: $0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8$

Step-by-step explanation:

We know that the standard quadratic equation is ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

$b^2-4ac=(-7)^2-4(2)(-9)=49+72>0$

So it has 2 real number solutions.

2. a=1, b=-4, c=4

$b^2-4ac=(-4)^2-4(1)(4)=16-16=0$

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

$b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0$

So it has 2 real number solutions.

4. a=1, b=-2, c=-8

$b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0$

So it has 2 real number solutions.

5. a=3, b=5, c=3

$b^2-4ac=(5)^2-4(3)(3)=25-36=-9$

Thus it does not has real solutions.

5 0

3 years ago

Read 2 more answers