Find the median for 8, 11, 6, 14, x, 13, and 11

Angie and Becky each completed a separate proof to show that the measures of vertical angles AKG and HKB are equal. Who complete

Svetllana [295]

Vertical angles are angles which are opposite to each other and have equal measures. With respect to the given proof, Angie completed the proof incorrectly.

Two angles are said to be vertical when they are opposite to one another and have equal measures. However, two angles are supplementary if they add up to 180°.

So, comparing Angie's and Becky's proofs, it can be deduced that from Angies proof:

1. Segment GH intersects segment AB at K (1. Given)

2. m∠AKG + m∠GKB = 180° (2. Definition of Supplementary Angles)

m∠GKB + m∠HKB = 180° (2. Definition of Supplementary Angles)

3. m∠AKG + m∠ GKB = m∠GKB + m∠HKB (3. Substitution Property)

4. m∠AKG = m∠HKB (4. Subtraction Property)

Note:

Step 2 should be Angle Addition Postulate, since:

m∠AKG + m∠GKB = 180°

m∠GKB + m∠HKB = 180°

So that;

m∠AKG + m∠GKB = m∠GKB + m∠HKB

Subtract m∠GKB from both sides to have;

m∠AKG = m∠HKB (which is the proof for step 4)

Therefore, Angie completed the poof incorrectly because of the properties of steps 2 and 4.

A diagram is herewith attached for more clarifications.

Visit: brainly.com/question/24529637

8 0

2 years ago

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

3 years ago

Graph the function f(x) = - squared x + 2

MariettaO [177]

One of the ways to graph this is to use plug in a few x-values and get an idea of the shape. Since the x values keep getting squared, there is an exponential increase on either side of the y-axis. You can see this by plugging in a few values:

When

x=0,f(x)=0

x=1,f(x)=1^2=1

x=2,f(x)=2^2=4

x=3,f(x)=3^2=9

x=4,f(x)=4^2=16

The same holds true for negative x-values to the left of the y-axis since a negative value squared is positive. For example,

x=−1,f(x)=(−1)2=1*−1=1

x=2,f(x)=(−2)2=−2*−2=4

The graph of f(x)=x^2 is called a "Parabola." It looks like this:

5 0

2 years ago

(x+3)²(65)=x(6) Solve using quadric formula

Angelina_Jolie [31]

Step 1: Simplify both sides of the equation.

65x²+390x+585=6x

Step 2: Subtract 6x from both sides.

65x²+390x+585−6x=6x−6x
65x²+384x+585=0

For this equation: a=65, b=384, c=585

65x²+384x+585=0

Step 3: Use quadratic formula with a=65, b=384, c=585

x=−b±√b2−4ac/2a
x=−(384)±√(384)2−4(65)(585)/
2(65)
x=−384±√−4644/130

Therefore, There are no real solutions.

6 0

2 years ago

Read 2 more answers

What is the sum of 1 1/2 and 2/3

prisoha [69]

$1\dfrac{1}{2}+\dfrac{2}{3}=1+\dfrac{1\cdot3}{2\cdot3}+\dfrac{2\cdot2}{3\cdot2}=1+\dfrac{3}{6}+\dfrac{4}{6}=1+\dfrac{3+4}{6}=1+\dfrac{7}{6}=1+1\dfrac{1}{6}=2\dfrac{1}{6}\\\\Answer:\ \boxed{1\dfrac{1}{2}+\dfrac{2}{3}=2\dfrac{1}{6}}$

6 0

3 years ago

Read 2 more answers