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butalik [34]
3 years ago
15

What slope would a line have to be perpendicular to a line with a slope of 1/4 ?

Mathematics
2 answers:
denpristay [2]3 years ago
8 0

Answer:

A line perpendicular to another has a slope that is the negative reciprocal of the slope of the other line. The negative reciprocal of the original line is –2, and is thus the slope of its perpendicular line.

Step-by-step explanation:

AURORKA [14]3 years ago
7 0

Answer:

The slope is -4

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9x⁴+6x²...................​
Lubov Fominskaja [6]

Answer:

9x^4=6,561x

6x^2=36x

6561x+36x=6597x

Step-by-step explanation:

9x to the 4th power is just 9x9=81x9 and so on and the same goes for 6 then you just add them. Not Sure if this is the way you needed but feel free to ask anything.

3 0
3 years ago
Read 2 more answers
I'm in algebra please help and explain with work
lidiya [134]
2x+4y=0
3x+y=10
So what we want to do is get y onto one side by itself using whatever equation you like.
So I'm taking 2x+4y=0 and subtracting 4y from both sides
So we end up with
2x=4y
Divide both sides by four
2/4x is 1/2 x
So we get 1/2x=y
Take this and plug it into the other equation
3x+y=10
3x+1/2x=10
Add your like terms
3 1/2x=10
Putting 3 1/2 as a mixed number we get 7/2
So..
7/2x=10
Multiply by the opposite of 7/2 on both sides
Which is 2/7 this will cancel out the left side.
So we end up with
X= 10•2/7
X=20/7
Which is 2 and 6/7 as a mixed number.

5 0
3 years ago
168 h = [?] days helppp
Nitella [24]

24 h in 1 day

just divide the hours by 24

168 : 24 = 7

6 0
3 years ago
I need the Combination( probability) of this problem.
coldgirl [10]
Check attached image file.

8 0
3 years ago
Will give brainliest!
Gala2k [10]
Check the picture below, it hits the ground when y = 0.

\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
t=\cfrac{162}{16}\implies t=\cfrac{81}{8}\implies t=10\frac{1}{8}

6 0
3 years ago
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