All categories

3 years ago

Given that AD is the perpendicular bisector of BC, AB=15, AC=x, and BD=0.25x, identify BC.

Mathematics

2 answers:

Colt1911 [192]3 years ago

6 0

Answer:

7.5 units

Step-by-step explanation:

BD = DC

Triangles ABD and ACD are congruent.

So, x = 15

Reasoning (if needed)

AC² = AD² + DC²

x² = AD² + (0.25x)²

AD² = 0.9375x²

AB² = AD² + DB²

15² = 0.9375x² + (0.25x)²

225 = x²

x = 15

BC = 2BD

= 2(0.25×15)

= 7.5

zvonat [6]3 years ago

3 0

Answer:

The answer is 7.5

Step-by-step explanation:

You might be interested in

2x+4y=16 what is the slope intercept form

Eva8 [605]

Since slope intercept is y=mx+b, you first want to get rid of the 2x by subtracting it from both sides. after, you should end up with 4y=16-2x or 4y=-2x+16, they both are the same, but some teachers might prefer the slope first. after that, you want to get the y alone, so you divide everything by 4. you should end up with the slope intercept form being y=-2/4x+4, simplified to y=-1/2x+4.

3 0

3 years ago

What does (r−c)(3) mean about George's new store?

ANTONII [103]

subtract r from c to get 8x+10

replace x with 3

8(3) +10 = 34 which in hundreds is 3400 dollars

the answer is C

3 0

2 years ago

Radwyn Horticulture is a company which focuses upon selling saplings. The company can grow a sapling for $15.45, and can sell th

SOVA2 [1]

Its A on Edge. Good Luck!

8 0

3 years ago

Read 2 more answers

The following diagram shows part of the graph of the function f(x)= -12x + 26 and g(x) = -(1/5) ^ -x + 2 + 3 what are the soluti

Nina [5.8K]

The answer would be D. For you to find the solution you would need to look at the x from the coordinates.

6 0

2 years ago

f(x) = square root ( x^4 -16x^2) let f be the function given by: . . a) Find the domain of f.. b) Find f '(x). c) Find the slope

Bond [772]

A ) The domain:
$x^{4} -16 x^{2} \geq 0 \\ x^{2} ( x^{2} -16) \geq 0 \\ x^{2} -16 \geq 0 \\ x^{2} \geq 16$
x ≤ -4 and x≥ 4 or: x ∈ ( - ∞ , - 4 ) ∪ ( 4 , + ∞ )
b ) f ` ( x ) = $\frac{1}{2 \sqrt{ x^{4} -16 x^{2} } } *( 4 x^{3} - 32 x ) = \\ = \frac{2x( x^{2} -8)}{ \sqrt{ x^{4}-16 x^{2} } }$
c ) The slope of the line normal to the graph at x = 5
m = -1 / f `(x)
f ` ( x ) = $\frac{10(25-8)}{ \sqrt{625-400} }= \frac{10*17}{15} = \frac{34}{3}$
m = - 3/34

8 0

3 years ago