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2 years ago

Find AB, given that line AD is the perpendicular bisector of BC

Mathematics

1 answer:

astraxan [27]2 years ago

5 0

Answer:

AB = 9

Step-by-step explanation:

The perpendicular bisector divides Δ ABC into 2 congruent triangles, that is

Δ ABD ≅ Δ ACD , then

AB = AC = 9

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There are 450 computer's in a technology store. Alex Bought 231 of them for school. Each one was $230. He has $10,000. How much

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2 years ago

What is the wavelength of a 320 Hz sound traveling 330 m/s?

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2 years ago

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2 years ago

H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.

natima [27]

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

<h3>Given information-</h3>

The given function is,

$h(x)=x^2+1$

$k(x)=x-2$

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

$(h+k)(x)=h(x)+k(x)$

$(h+k)(x)=x^2+1+x-2$

$(h+k)(2)=2^2+1+2-2$

$(h+k)(x)=5$

2)The value of the function (h-k)(3),

$(h-k)(x)=h(x)-k(x)$

$(h-k)(x)=x^2+1-x+2$

$(h-k)(3)=3^2+1-3+2$

$(h-k)(x)=9$

3) The value of the function 3h(2)+2k(3)

$3h(x)+2k(x)=3x^2+3+2x-2\times 2$

$3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2$

$3h(2)+2k(3)=17$

Hence the value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

Learn more about the quadratic equation here;

brainly.com/question/2263981

4 0

2 years ago