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iris [78.8K]
3 years ago
14

Help me solve these two questions please!

Mathematics
1 answer:
ozzi3 years ago
5 0
They are simple linear equations with one unknown, lets tackle them, one step at the time solving for the unknown:
4x - 2(x - 5) = x + 13
4x - 2x + 10 = x + 13
2x + 10 = x + 13
2x - x = 13 - 10
x = 3
that is the solution.

3(6 - x) - 4 = 5x + 2(7x + 3)
18 - 3x - 4 = 5x + 14x + 6
14 - 3x = 19x + 6
14 -  6 = 19x + 3x
8 = 22x
x = 8/22 = 4/11
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A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 1.5 feet, 30 feet a
Nimfa-mama [501]

Answer:

x₁ > x₂

Step-by-step explanation:

Both actions imply a parable trajectories, since both are projectile shot cases.

Let´s call x₁ maximum distance in the first case

The maximum height is just in the middle of the curve, therefore x₁ the maximum horizontal distance is equal to 60 feet.

In the second case, the parable curve is modeled by:

y = x₂*( 0.08 - 0.002x₂)    or  y = 0.08*x₂ - 0.002*x₂²

A second degree equation, solving for x₂ and dismissing the value x₂ = 0

we get:

y = 0       ⇒    x₂*( 0.08 - 0.002x₂) = 0    x₂ = 0

And 0,08 - 0.002*x₂ = 0

- 0.002*x₂ = - 0.08

x₂ = 0.08/0.002

x₂ = 40 f

Then  x₁ > x₂

3 0
3 years ago
Need help <br> Please<br> Thank u
Volgvan

Answer:

i think its b

Step-by-step explanation:

4 0
3 years ago
Can someone please please pleaseee helppp
uranmaximum [27]
My lucky guess would be 3
4 0
3 years ago
Read 2 more answers
A bicycle tire is 28 inches in diameter. Approximately how far does the bicycle move forward each time the wheel goes around? (U
boyakko [2]

Answer:

Step-by-step explanation:

28x pi=87.96 in

8 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

3 0
2 years ago
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