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emmasim [6.3K]
3 years ago
7

What is the measure of

Mathematics
1 answer:
bagirrra123 [75]3 years ago
8 0
The answer is A. 74°
If the vertex of an angle is on the circle, then the angle measure is one-half the intercepted arc.
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PLS 40 points
Pavlova-9 [17]
<h2>2x+y=2</h2>

Step-by-step explanation:

Let p1 be the point (-1,4)

Let p2 be the point (3,-4)

The equation of the line passing through two points p1=(x_{1},y_{1})

and p2=(x_{2},y_{2}) is \frac{y-y_{1}}{x-x_{1}} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

substituting p1,p2 in the above equation yields

\frac{y-4}{x-(-1)}=\frac{-4-4}{3-(-1)}

which when simplified gives \frac{y-4}{x+1}=\frac{-8}{4}

which when further simplified gives 2x+y=2

5 0
3 years ago
Find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. ​
borishaifa [10]

Answer:

To find the surface area of a pyramid, start by multiplying the perimeter of the pyramid by its slant height. Then, divide that number by 2. Finally, add the number you get to the area of the pyramid's base to find the surface area.

Step-by-step explanation:

7 0
3 years ago
Please help me with these questions
Citrus2011 [14]

Answer:

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8 0
3 years ago
Solve for x<br> (3x + 2)<br> 71<br> A) 28<br> C) 24<br> B) 23<br> D) 34
Sloan [31]

3x=71-2

3x=69

x=69/3

x=23

7 0
4 years ago
Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

4=(k^2-6k)+(k-2)

4=k^2-5k-2

0=k^2-5k-6

We can factorize the right side:

0=(k-6)(k+1)

which tells us that k=6 and k=-1 are solutions.

4 0
4 years ago
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