Answer:
- <em>To solve these first swap x and y, solve for y and name it inverse function</em>
3. <u>y = -8x + 2</u>
- x = -8y + 2
- 8y = -x + 2
- y = -x/8 + 2/8
- y = -(18)x + 1/4
f⁻¹(x) = -(18)x + 1/4
-----------------------------------------
4.<u> y = (2/3)x - 5</u>
- x = (2/3)y - 5
- (2/3)y = x + 5
- y = (3/2)x + (3/2)5
- y = 1.5x + 7.5
f⁻¹(x) = 1.5x + 7.5
-----------------------------------------
5. <u>f(x) = 2x² - 6</u>
- x = 2y² - 6
- 2y² = x + 6
- y² = 1/2x + 3
- y =
f⁻¹(x) =
-----------------------------------------
6. <u>y = (x - 3)²</u>
- x = (y - 3)²
= y - 3- y = 3 +
f⁻¹(x) = 3 +
Answer:
Kaylee drove her scooter for 2h and Sophia drove her scooter for 3h.
Step-by-step explanation:
From the graph shown below, we can see that the intersection point is (2,3). When using the graphing method to solve a system of equations, the intersection point is the answer. In this case, x = 2 and y = 3.
Thus, Kaylee drove her scooter for 2h and Sophia drove her scooter for 3h.
Plotting the data (attached photo) roughly shows that the data is skewed to the left. In other words, data is skewed negatively and that the long tail will be on the negative side of the peak.
In such a scenario, interquartile range is normally the best measure to compare variations of data.
Therefore, the last option is the best for the data provided.
Answer:
z<24
Step-by-step explanation:
It can't be more than one value. All you do is subtract 35 from both sides to isolate z.
Hope this helped!
let's first off take a peek at those values.
let's say the point with those coordinates is point C, so C is 3/10 of the way from A to B.
meaning, we take the segment AB and cut it in 10 equal pieces, AC takes 3 pieces, and CB takes 7 pieces, namely AC and CB are at a 3:7 ratio.
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-4,-8)\qquad B(11,7)\qquad \qquad \stackrel{\textit{ratio from A to B}}{3:7} \\\\\\ \cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em] ~\dotfill\\\\ C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%0A%5C%5C%5C%5C%5C%5C%0AA%28-4%2C-8%29%5Cqquad%20B%2811%2C7%29%5Cqquad%0A%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20B%7D%7D%7B3%3A7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BA%5Cunderline%7BC%7D%7D%7B%5Cunderline%7BC%7D%20B%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BB%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%207A%3D3B%5Cimplies%207%28-4%2C-8%29%3D3%2811%2C7%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AC%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right) \\\\\\ C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20C%3D%5Cleft%28%5Ccfrac%7B%287%5Ccdot%20-4%29%2B%283%5Ccdot%2011%29%7D%7B3%2B7%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%287%5Ccdot%20-8%29%2B%283%5Ccdot%207%29%7D%7B3%2B7%7D%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AC%3D%5Cleft%28%20%5Ccfrac%7B-28%2B33%7D%7B10%7D~~%2C~~%5Ccfrac%7B-56%2B21%7D%7B10%7D%20%5Cright%29%5Cimplies%20C%3D%5Cleft%28%20%5Ccfrac%7B5%7D%7B10%7D~~%2C~~%5Ccfrac%7B-35%7D%7B10%7D%20%5Cright%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20C%3D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D~%2C~-%5Cfrac%7B7%7D%7B2%7D%20%5Cright%29~%5Chfill)
