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3 years ago

Which postulate or theorem proves that triangle JHG and

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

The answer is AS A Congruence Postulate.

Step-by-step explanation:

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Represent A Quadratic Function Using:

Pani-rosa [81]

Sub the Xs in the eqn

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2 years ago

Please help with this

klemol [59]

Answer:

(h+k)(2)=4+1+2-2=5

(h-k)(3)=9+1-3-2=3+1-1=9

3h2+2k3=3*5+2=17

7 0

3 years ago

Factor out the combine term 21x+49y

Cloud [144]

(not sure if this is right just trying to help) 7(3x+7y)

7 0

3 years ago

Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)

timama [110]

<h3>Answer: c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

$f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\$

With a = -5 and b = 4, we have

$f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\$

So,

$f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\$

Use algebra to solve for c

$-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\$

6 0

3 years ago

Consider the point X at (0,21) and point Y at (88, -103). What is the coordinate of the midpoint of xy

saul85 [17]

Step-by-step explanation:

let M be the midpoint of x and y

xM= (xX+xY)/2 =(0+88)2=44

yM=(Yx+yY)/2=(21_103)/2=_41

M(44,_41)

7 0

3 years ago