answer:To find the common solutions for these three inequalities, let's first solve each inequality individually. In (1): -3y geq y+7 Adding y to both sides gives: -2y geq 7 Dividing both sides by -2 gives: y leq -3.5 In (2): -2y leq 12 Dividing both sides by -2 gives: y geq -6 In (3): -4y geq 2y+17 Subtracting 2y from both sides gives: -6y geq 17 Dividing both sides by -6 gives: y leq -frac{17}{6} Now, let's find the region of overlap. From (1), we know that y leq -3.5. From (2), we know that y geq -6. From (3), we know that y leq -frac{17}{6}. The range where all three inequalities are satisfied is -frac{17}{6} leq y leq -3.5. There are infinitely many integers that are common solutions to these three inequalities because the common solution range includes all negative integers between -5 and -3, inclusive, and some additional non-integer values as well. The integers within this range are: ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, ..., First, we need to solve each inequality: (1) -3y geq y + 7 Subtract y from both sides: -4y geq 7 Divide by -4 and reverse the inequality sign: y leq -frac{7}{4} (2) -2y leq 12 Divide by -2 and reverse the inequality sign: y geq -6 (3) -4y geq 2y + 17 Subtract 2y from both sides: -6y geq 17 Divide by -6 and reverse the inequality sign: y leq -frac{17}{6} The common solution to all three inequalities is the intersection of the solution sets. Since y leq -frac{7}{4} and y geq -6, the common interval is -frac{7}{4} leq y leq -6. However, there is no integer value within this interval. Therefore, there are no integer solutions common to all three inequalities.,To find the common solutions, I need to find the intersection of the regions defined by each inequality on the number line. I can do this by solving each inequality for y and finding the endpoints of the corresponding intervals. For inequality (1), I can add 3y to both sides and subtract 7 from both sides to get 4y leq -7. Then I can divide both sides by 4 to get y leq -7/4. This means that any integer y that is less than or equal to -7/4 is a solution to (1). For inequality (2), I can divide both sides by -2, but I have to remember to flip the inequality sign when I do that, so I get y geq -6. This means that any integer y that is greater than or equal to -6 is a solution to (2). For inequality (3), I can add 4y to both sides and subtract 17 from both sides to get 6y leq -17. Then I can divide both sides by 6 to get y leq -17/6. This means that any integer y that is less than or equal to -17/6 is a solution to (3). Now I have three intervals on the number line: y leq -7/4, y geq -6, and y leq -17/6. To find the intersection of these intervals, I need to find the largest lower bound and the smallest upper bound among these endpoints. The largest lower bound is -6, since it is bigger than both -7/4 and -17/6. The smallest upper bound is -17/6, since it is smaller than both -7/4 and -6. Therefore, the intersection of the intervals is -6 leq y leq -17/6. However, I only want the integer solutions, so I need to round the endpoints to the nearest integers that are within the interval. The nearest integer to -6 that is within the interval is -6 itself. The nearest integer to -17/6 that is within the interval is -3, since -17/6 is about -2.83 and -3 is closer than -2. Therefore, the integer solutions are -6, -5, -4, and -3. There are four of them.

answer:To evaluate this expression, we'll first round off each intermediate result: Original expression: leftlfloor leftlceil left(frac{13}{7}right)^2rightrceil+frac{17}{4}rightrfloor Step 1: Calculate (frac{13}{7})^2 = (2.28571428571)2 approx 5.17636428571 Step 2: Round up (frac{13}{7})^2 to leftlceil 5.17636428571rightrceil = lceil 5.176rceil = 6 Step 3: Round lceil left(frac{13}{7}right)^2rceil to leftlceil left(frac{13}{7}right)^2rightrceil = ⌈⌉ = (6) Step 4: Calculate frac{17}{4} = 4.25 Step 5: Round up frac{17}{4} to leftlceil frac{17}{4}rightrceil =⌈4.25⌉ = 5 Step 6: Add leftlceil left(frac{13}{7}right)^2rightrceil and leftlceil frac{17}{4}rightrceil = 6 + 5 = 11 Step 7: Round leftlceil 5.176rceil + leftlceil 4.25rightrceil to lfloor 11 rfloor = ⌊⌋ = 11 And thus, the final expression is leftlfloor leftlceil left(frac{13}{7}right)^2rightrceil+frac{17}{4}rightrfloor = 11., Let's evaluate the expression from the inside out. First, calculate left(frac{13}{7}right)^2: left(frac{13}{7}right)^2 approx 2.9605 Next, apply the ceiling function: leftlceil 2.9605 rightrceil = 3 Now, add frac{17}{4}: 3 + frac{17}{4} = frac{27}{4} + frac{17}{4} = frac{44}{4} = 11 Finally, apply the floor function: leftlfloor 11 rightrfloor = 11 So, the result of leftlfloor leftlceil left(frac{13}{7}right)^2rightrceil+frac{17}{4}rightrfloor is 11.,This problem involves applying the floor and ceiling functions to some fractions and exponents. The floor function, denoted by lfloor x rfloor, rounds x down to the nearest integer. The ceiling function, denoted by lceil x rceil, rounds x up to the nearest integer. To evaluate this expression, I need to work from the inside out, applying the functions in the correct order. First, I will evaluate the innermost expression, which is left(frac{13}{7}right)^2. To do this, I can multiply the numerator and denominator by themselves: left(frac{13}{7}right)^2=frac{169}{49}. Next, I will apply the ceiling function to this fraction. To do this, I need to round it up to the nearest integer. Since frac{169}{49} is between 3 and 4, the nearest integer that is greater than or equal to it is 4. So, lceil left(frac{13}{7}right)^2 rceil = 4. Now, I will evaluate the next expression, which is lceil left(frac{13}{7}right)^2 rceil+frac{17}{4}. To do this, I can substitute 4 for lceil left(frac{13}{7}right)^2 rceil and add the fractions. To add the fractions, I need a common denominator, which is 4. So, lceil left(frac{13}{7}right)^2 rceil+frac{17}{4}=frac{4}{1}+frac{17}{4}=frac{16}{4}+frac{17}{4}=frac{33}{4}. Finally, I will apply the floor function to this fraction. To do this, I need to round it down to the nearest integer. Since frac{33}{4} is between 8 and 9, the nearest integer that is less than or equal to it is 8. So, lfloor lceil left(frac{13}{7}right)^2 rceil+frac{17}{4} rfloor = 8.

answer:The slope of a line can be found by rearranging the equation into the form y=mx+b. To find the slope of a line perpendicular to this line, we must first find the slope of the initial line. First, we'll rearrange the given equation to have the y term on the left side: 3y = -2x + 6 Then, we can solve for y: y = frac{-2x + 6}{3} The slope of the line (m) is the coefficient of x in the equation y=mx+b. So, the slope in this case is: m = -2 Now, we can find the slope of a line perpendicular to this one. Perpendicular lines have slopes that are negative reciprocals of one another. So, the slope of the perpendicular line would be the reciprocal of -2 which is frac{1}{2}. In conclusion, the slope of the line perpendicular to 2x + 3y = 6 is frac{1}{2}.,I know that two lines are perpendicular if their slopes are negative reciprocals of each other. That means I need to find the slope of the given line first, and then flip it over and change its sign. To find the slope of the given line, I can rearrange the equation into slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. Now I can see that the slope of the given line is -frac{2}{3}. That means I flip the fraction over and change the sign, so I get frac{3}{2}., To find the slope of a line perpendicular to a given line, we first need to find the slope of the given line. The equation of the given line is in the form ax + by = c, so we can rewrite it in slope-intercept form (y = mx + b) to find the slope. The given equation is 2x + 3y = 6. To put this in slope-intercept form, we'll solve for y: 3y = -2x + 6 y = -frac{2}{3}x + 2 Now we can see that the slope of the given line is m = -frac{2}{3}. The slope of a line perpendicular to a given line with slope m is the negative reciprocal of m, or m_perp = -frac{1}{m}. Therefore, the slope of the line perpendicular to our given line is: m_perp = -frac{1}{-frac{2}{3}} = boxed{frac{3}{2}}

answer:I see that both the numerator and the denominator have the same number of digits, and they are all either 2's or 6's. This makes me think that there might be a way to simplify the fraction by canceling out some common factors. One thing I notice is that 66,666 is three times 22,222, so I can write the fraction as displaystyle frac{(3 cdot 22,!222)^4}{22,!222^4}. Now I can use the exponent rule (a cdot b)^n = a^n cdot b^n to expand the numerator: displaystyle frac{(3 cdot 22,!222)^4}{22,!222^4} = frac{3^4 cdot (22,!222)^4}{(22,!222)^4}. Now I see that I can cancel out the (22,!222)^4 terms in the numerator and the denominator, leaving me with displaystyle frac{3^4 cdot (22,!222)^4}{(22,!222)^4} = 3^4. Finally, I can evaluate 3^4 = 3 cdot 3 cdot 3 cdot 3 = 9 cdot 9 = 81.,To compute this expression, you can use the properties of exponents. dfrac{66,!666^4}{22,!222^4} = dfrac{(66,!666^2)^2}{(22,!222^2)^2} Next, notice that both the base 66,666 and 22,222 in the numerator and denominator are very similar. You can simplify the fraction further by dividing each part by the common factor: dfrac{(66,!666^2)^2}{(22,!222^2)^2} = dfrac{66,!666^2 div 22,!222^2}{cancel{22,!222^2} div cancel{22,!222^2}} = dfrac{66,!666^2}{22,!222^2} Now, you can use the property that (a^m)^n = a^{mn} to further simplify the expression: dfrac{66,!666^2}{22,!222^2} = dfrac{(66,!666 cdot 66,!666)}{(22,!222 cdot 22,!222)} = dfrac{66,!666 cdot 66,!666}{22,!222 cdot 22,!222} Finally, multiply the numbers in the fraction: dfrac{66,!666 cdot 66,!666}{22,!222 cdot 22,!222} = dfrac{440,!184 cdot 440,!184}{440,!184} = 440,!184, We have: displaystyle frac{66,!666^4}{22,!222^4} = left( frac{66,!666}{22,!222} right)^4 = left( frac{3}{1} right)^4 = 81. Therefore, displaystyle frac{66,!666^4}{22,!222^4} = 81.