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DiKsa [7]
3 years ago
5

Which of the following describes a linear function

Mathematics
1 answer:
liq [111]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Linear function means that the y values depend upon the x values and any other numbers that are being added to it.

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Is 144 a perfect square or not​
Rina8888 [55]

Answer: Yes

Step-by-step explanation: 144 is a perfect square. In other words, it's possible to find a whole number that can be multiplied by itself that gives us 144.

In this case, we multiply 12 by 12 or 12² to get 144 which means 144 is a perfect square. Also, it's on our perfect square list which I will attach below.

6 0
3 years ago
Linear programming: You're making fruit baskets. Each day you have 240 oranges, 270 bananas, and 320 apples. Arrangement A earns
8_murik_8 [283]

Answer:

知りません

Step-by-step explanation:

5 0
3 years ago
Evaluate x/y + 5 w^2, for x = 2/3 and y = 5/6, and w = 3
tresset_1 [31]
2/3 / 5/6 + 5 (3)^2

= 2/3 * 6/5 + 45

= 12/15 + 45  

= 45 4/5  Answer
5 0
3 years ago
Solve the equation for x.
DiKsa [7]
You multiply 0.6 by 5x and 10 and you get 3x+6=-24. you then subtract 6 from both sides of the equation and you get 3x=-30. after that you divide both sides by 3 and you get x=-10. so the answer is B!
6 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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