MATHEMATICS HIGH SCHOOL

Point O is the center of the circle in the diagram. What is
m∠BCA?

Answers

Answer 1

Answer:

we know that

The measure of the external angle is the semidifference of the arcs that it covers.

$m\ angle\ BCA= (1)/(2) *(arc\ BOA-arc\ BA)\n\n m\ angle\ BCA= (1)/(2) *(250-110)\n \n m\ angle\ BCA=70\ degrees$

therefore

the answer is

The measure of angle BCA is equal to $70\ degrees$

Answer 2

Answer: As you know the exterior angle formed by m< BOA is 250 degrees.

The point where C is the point of vertical angles. Vertical angles form a linear pair.
Linear pair is Always of 180 degrees so,
250 - 180 = 70 degrees

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What is 2x + 18 = 20?????

Complete the statement, then write the postulate you used. Please help! I’ll give brainliest!

Answers

The answer is MN with a bar/line on top of it

Which expression is undefined? A.0 ÷ 5

B.(3 - 3)
_____
7

C. 8
_____
(-2+2)

D. (0 - 12)
_____
-6

Answers

Answer: C

Work: So, after you simplify that equation, it's 8/0. You Cannot divide zero like that. You can divide 0/8, but you can't with 8/0. I hope this wasn't to confusing, and Happy Holidays! :)

Find the solution of 3 times the square root of the quantity of x plus 5 equals negative 9, and determine if it is an extraneous solution.

Answers

Answer: x= 4 is an extraneous solution.

Step-by-step explanation:

Since we have given that

$3(√(x+5))=-9$

We need to find the extraneous solution.

So, our equation becomes,

$√(x+5)=(-9)/(3)\n\n√(x+5)=-3\n\n\text{On squaring both sides}\n\nx+55=(-3)^2\n\nx+5=9\n\nx=9-5\n\nx=4$

Now, we will check that x = 4 is an extraneous solution.

$3√(4+5)\neq -9\n\n3√(9)\neq -9\n\n3* 3\neq -9\n\n9\neq -9$

Hence, x= 4 is an extraneous solution.

The solution is $x=4$ .

The seventh term of the sequence -2, -6, -10, -14, ... is _____. -26 -24 -22 -20

Answers

We'll it keeps on subtracting by 4 to get each number.

-2,-6,-10,-14,-18,-22,-26.

Negative 26 is your answer. Hope this helped!

Answer:

-26

Step-by-step explanation:

We are counting by 4s

1) 2 - 4 = -2

2) -2 - 4 = -6

3) -6 - 4 = -10

4) -10 - 4 = -14

5) -14 - 4 = -18

6) -18 - 4 = -22

7) -22 - 4 = -26

Hope this helps!!!

Solve: y=6x-8 and y=-3x+10

Answers

The solution to the system of equations is x = 2 and y = 4.

To solve the system of equations:

y = 6x - 8 ...(Equation 1)

y = -3x + 10 ...(Equation 2)

We can set the two equations equal to each other:

6x - 8 = -3x + 10

To solve for x,

6x + 3x = 10 + 8

9x = 18

Dividing both sides by 9:

x = 18/9

x = 2

So, y = 6(2) - 8

y = 12 - 8

y = 4

Therefore, the solution to the system of equations is x = 2 and y = 4.

Learn more about Equation here:

brainly.com/question/29657983

#SPJ6

Answer:

x = 2

Step-by-step explanation:

hey there,

< If these two are together, then the problem would look like this:

$\left \{ {{y=6x-8} \atop {y=-3x+10}} \right.$

From the first equation, we can see that y = 6x-8. I am assuming in your problem you need to find what "x" is equal to, so plug in the first "y" value into the second one.

6x - 8 = -3x + 10

Bring all "x"s to one side and regular numbers to the other side.

6x + 3x = 10 + 8

9x = 18

x = 2

x = 2 is your final answer. >

Hope this helped! Feel free to ask anything else.

The half-life of Po-214 is 0.001 seconds. How much of a 10g sample is left after the 0.003 seconds. Also, create an exponential function to model its decay.

Answers

$\bf \textit{Amount for Exponential Decay using Half-Life} \n\n A=P\left( (1)/(2) \right)^{(t)/(h)}\qquad \begin{cases} A=\textit{accumulated amount}\n P=\textit{initial amount}\dotfill &10\n t=\textit{elapsed time}\dotfill &0.003\n h=\textit{half-life}\dotfill &0.001 \end{cases} \n\n\n A=10\left( (1)/(2) \right)^{(0.003)/(0.001)}\implies A=10\left( (1)/(2) \right)^3\implies A=1.25$

Final answer:

After 0.003 seconds, we would have 1.25g of the initial 10g Po-214 left. The decay process can be modeled with the exponential function: N = 10 * $(1/2)^{(0.003/0.001)$ , where N is the final amount, N0 is the initial amount, t is time, and T is the half-life.

Explanation:

The subject matter pertains to the concept of half-life in physics which interprets the decay of a radioactive material. In radioactive decay, the number of atoms in a radioactive sample decreases exponentially over time. This time is taken into consideration in the form of their half-lives. In the case of Po-214, the half-life is given as 0.001 seconds.

Before finding how much of Po-214 is left after 0.003 seconds, let's first understand the calculation. After one half-life (0.001 second), half of the sample will be left, i.e., 10g/2 = 5g. After another half-life (another 0.001 second), half of this 5g will be left, i.e.,5g/2 = 2.5g. After the third half-life (the remaining 0.001 second), we will have 2.5g/2 = 1.25g left. So, after 0.003 seconds, we would have 1.25g of Po-214 remaining from the initial 10g.

The decay process can be modeled with an exponential function N = N0 * $(1/2)^{(t/T)$ , where N is the final amount, N0 is the initial amount, t is time, and T is the half-life. For this problem, the equation would be: N = 10 * (1/2)^(0.003/0.001).