Answer:
I think its the 1st one, but I'm not sure
Answer:
A and D
Step-by-step explanation:
67 right out of 100 for the fraction and 67/100 = 67%
<span>Angle TSQ measures 68 degrees.
When a ray bisects an angle, it divides it into two equal parts. Each part is one-half the measurement of the original angle. Several rays are described as bisecting different angles. I would sketch a diagram to keep track of all the different rays and angles.
A. Since angle RST is bisected by ray SQ, angle RSQ and angle QST are each half the size of angle RST.
B. Since angle RSQ is bisected by ray SP, angle RSP and angle PSQ are each half the size of angle RSQ.
C. Since angle RSP is bisected by ray SV, angle RSV and angle VSP are each half the size of angle RSP.
We are given the measurement of angle VSP as 17 degrees. To find the measure of angle RSP, we notice in statement C above that VSP is half the size of angle RSP. If we double angle VSP's measurement (multiply by 2), we get angle RSP measures 34 degrees.
Using similar logic and statement B above, we double RSP's measurement of 34 to get angle RSQ's measurement. Double 34 is 68, angle RSQ's measurement in degrees.
From statement A above, we notice that RSQ's measurement is equal to that of angle QST's. Therefore, angle QST also measures 68 degrees. However, the question asks us to find the measurement of angle TSQ. However, angle QST and angle TSQ are the same. Either description can be used. Therefore, the measurement of angle TSQ is 68 degrees.</span>
Answer:
The equation does not have a real root in the interval ![\rm [0,1]](https://tex.z-dn.net/?f=%5Crm%20%5B0%2C1%5D)
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if 
is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval , which means to evaluate the equation in 0 and 1:
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval . I attached a plot of the equation in the interval ![\rm [-2,2]](https://tex.z-dn.net/?f=%5Crm%20%5B-2%2C2%5D)
where you can clearly observe how the graph does not cross the x-axis in the interval.
