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4 years ago

the expression 4(1.5g+ 3.5) is equal to 6g plus a constant term what is the value of the constant term

Mathematics

1 answer:

riadik2000 [5.3K]4 years ago

3 0

4*3.5 =14.0
......................

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Which expression gives the distance between the points (4,6) and (7,-3)?

Stella [2.4K]

Answer: The distance between is 13

Step-by-step explanation:

The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √ ((X2-X1)2+(Y2-Y1)2)

d = √ (4-4)2+(-6-7)2

d = √ ((0)2+(-13)2)

d = √ (0+169)

d = √ 169

The distance between the points is 13

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3 years ago

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Likurg_2 [28]

The answer is c have a wonderful day

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3 years ago

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irina [24]

Answer:

0.95

Step-by-step explanation:

19/20 as a decimal is 0.95

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3 years ago

Read 2 more answers

The cost of tickets to a circus is expressed by the function y = 4x + 2, where

-BARSIC- [3]

Answer:

Step-by-step explanation:

This is a linear function, written in the form y=mx+b, where m is the slope, or how many dollars each ticket costs in the case, and b is the initial value of the function, or in this case the cost before any tickets are added to the account.

Since 4x is the slope (4 times however many tickets one purchased), 2 is the initial value before we buy tickets.

You can also plug in 0 for x and get $4(0)+2=2$

Hope I helped :D

6 0

3 years ago

Uestion

Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}$

now, let's check the other path, JM and KL

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}$

so the red path will be $5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare$

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2 years ago