Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
You use the quadratic formula:
2x {}^{2} - 3x = 5
2x {}^{2} - 3x - 5 = 0
x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2}
x = \frac{3 + - \sqrt{49} }{4}
x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4}
x = \frac{5}{2} \: and \: x = - 1
Steps:
1. since the least common multiple of 8 and 4 is 8, you would have to make both of the denominators 8.
2. 2 times 4 equals 8, 2 times 3 equals 6
3. 35 3/4 now becomes 35 6/8
4. your new problem is 35 6/8 - 1/8
5. 6 -1= 5
6. your answer is 35 5/8
Answer:
6: compression 7: translation 8: Reflection
Answer:
the anwser is y=5600
I'm going to just go by this because of research and you didn't show the 4 problems so if this fixes your question then there you go
